Answer:
0.0583g
Explanation:
The equation of the reaction is;
2HNO3(aq) + Mg(OH)2(aq) -------> Mg(NO3)2(aq) + 2H2O(l)
From the question, number of moles of HNO3 reacted= concentration × volume
Concentration of HNO3= 0.100 M
Volume of HNO3 = 20.00mL
Number of moles of HNO3= 0.100 × 20/1000
Number of moles of HNO3 = 2×10^-3 moles
From the reaction equation;
2 moles of HNO3 reacts with 1 mole of Mg(OH)2
2×10^-3 moles reacts with 2×10^-3 moles ×1/2 = 1 ×10^-3 moles of Mg(OH)2
But
n= m/M
Where;
n= number of moles of Mg(OH)2
m= mass of Mg(OH)2
M= molar mass of Mg(OH)2
m= n×M
m= 1×10^-3 moles × 58.3 gmol-1
m = 0.0583g
g Which ONE of the following pure substances will exhibit hydrogen bonding? A) methyl fluoride, FCH3 B) dimethyl ether, CH3C–O–CH3 C) formaldehyde, H2C=O D) trimethylamine, N(CH3)3 E) hydrazine, H2N-NH2
Answer:
C) formaldehyde, H2C=O.
Explanation:
Hello,
In this case, given that the hydrogen bondings are known as partial intermolecular interactions between a lone pair on an electron rich donor atom, particularly oxygen, and the antibonding molecular orbital of a bond between hydrogen and a more electronegative atom or group. Thus, among the options, C) formaldehyde, H2C=O, will exhibit hydrogen bonding since the lone pair of electrons of the oxygen at the carbonyl group, are able to interact with hydrogen (in the form of water).
Best regards.
What must happen to uranium before it can be used as a fuel source?
Answer: Uranium enrichment. Uranium is used to fuel nuclear reactors; however, uranium must be enriched before it can be used as fuel. Enriching uranium increases the amount of uranium-235 (U235) that can sustain the nuclear reaction needed to release energy and produce electricity at a nuclear power plant.
Hello, I am a bit stuck on this. Could someone help?
4. Given that the enthalpy of reaction for a system at 298 K is -292 kJ/mol and the entropy for that system is 224 J/mol K, what's the free energy for the system?
A.-87,793 kJ
B.-358 kJ
C.-225 kJ
D. -66,751 kJ
Answer:
[tex]\Delta G=-359\frac{kJ}{mol}[/tex]
Explanation:
Hello,
In this case, we must remember that the Gibbs free energy is defined in terms of the enthalpy, temperature and entropy as shown below:
[tex]\Delta G=\Delta H -T\Delta S\\[/tex]
In such a way, for the given data, we obtain it, considering the conversion from J to kJ for the entropy in order to conserve the proper units:
[tex]\Delta G=-292\frac{kJ}{mol} -(298)(224\frac{J}{mol}*\frac{1kJ}{1000J} )\\\\\Delta G=-359\frac{kJ}{mol}[/tex]
Best regards.
Answer:
B- 358 kj
Explanation: I took the test
Which option is an isotope with 6 protons and 8 neutrons? (1 point)
O oxygen-6
o carbon-14
o oxygen-14
o carbon-8
Answer:
Carbon-14
Explanation:
It is because it have 6 protons and 8 neutrons which are added together to form an atomic mass of 14.
The mass number of an isotope is the sum of the number of protons and number of neutrons hence the correct answer is carbon-14.
Isotopes are atoms of the same element that has the same number of protons but different number of neutrons.
The atomic number is the same as the number of protons in the atom. Since there are six protons in the atom, the element must be carbon.
There are eight neutrons meaning that the mass number is 8 + 6 = 14
Therefore, the isotope is carbon-14.
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Human blood typically contains 1.04 kg/L of platelets. A 1.89 pints of blood would contain what mass (in grams) of platelets
A 1.89 pints of blood would contain 873 grams of platelets.
To calculate the amount of platelets present in 1.89 pints, it is first necessary to transform this unit of volume into liters:
1 pint = 473.2 mL[tex]1.89 \times 473.2 = 894.3 mL[/tex]
1000 L = 1mL
[tex]\frac{894.3}{1000}= 0.84L[/tex]
Now, just calculate the amount of platelets present in 0.84L:
[tex]\frac{1.04\times10^{3}g}{xg}=\frac{1L}{0.84L}[/tex]
x = 873 grams
So, a 1.89 pints of blood would contain 873 grams of platelets.
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Razak uses 70 W fan and a 40 w lamp for eight hours a day, Calculate the amount of energy he used in a month. NORS Analysing
Answer:
70 W and 40 w
there are 28 days in a month
Add those 2 together and get 110
then mutlpy by 28
3,080 amount of energy
Fireworks are chemical reactions that release energy. Which of these phenomena are caused by chemical reactions that release energy? If you’re not sure, make a guess.
Answer:
All chemical reactions involve energy. Energy is used to break bonds in reactants, and energy is released when new bonds form in products. Endothermic reactions absorb energy, and exothermic reactions release energy. The law of conservation of energy states that matter cannot be created or destroyed.
what is the mass of cerrusite would contain 35g of lead?
Answer:
i believe its 42
Explanation:
The ph of the test tube can be calculated by knowing the concentration of hydroxide ions in solution. the ph = 14 + log 10 oh- for example the 0.1m stock of naoh has a ph = 14+ log 10 0.1= 13. using the dilution 5 ml 0.1 naoh 5ml h20 what is the ph of tube 1.
Answer:
Volume of the calcium hydroxide solution used is 0.0235 mL.
Explanation:
Moles of KHP =
According to reaction, 2 moles of KHP with 1 mole of calcium hydroxide , then 0.0330 moles of KHP will recat with ;
of calcium hydroxide
Molarity of the calcium hydroxide solution = 0.703 M
Volume of calcium hydroxide solution = V
Volume of the calcium hydroxide solution used is 0.0235 mL.
Given that the Ksp value for Ca3(PO4)2 is 8.6×10−19, if the concentration of Ca2+ in solution is 4.9×10−5 M, the concentration of PO3−4 must exceed _____ to generate a precipitate.
Answer:
.0027 M
Explanation:
We must calculate the threshold concentration of PO3−4 using Ksp and the given concentration of Ca2+:
Ca3(PO4)2(s)⇌3Ca2+(aq)+2PO3−4(aq)
Ksp=8.6×10−19=[Ca2+]3[PO3−4]2=(4.9×10−5M)3[PO3−4]2
[PO3−4]=0.0027 M
If the amount of radioactive iodine-123, used to treat thyroid cancer, in a sample decreases from 3.2 to 0.4 mg in 39.6 h, what is the half-life of iodine-123?
Answer:
Half life = 13.197 hour
Explanation:
Given:
Old amount (A₀) = 3.2
New amount (A) = 0.4
Radiation decay time (t) = 39.6 hour
Half life = T(1/2)
Find:
Half life = T(1/2) = T
Computation:
A = A₀[tex]e^{-(\frac{0.693t}{T} )}[/tex]
[tex]e^{-(\frac{0.693t}{T} )}[/tex] = 0.4 / 3.2
-[27.4428 / T] = In (0.125)
-[27.4428 / T] = -2.0794
[27.4428 / T] = 2.0794
T = 13.197
Half life = 13.197 hour
A 1.362 g sample of an iron ore that contained Fe3O4 was dissolved in acid with all of the iron being reduced to iron (II). The solution was acidified with sulfuric acid and titrated with 39.42 mL of 0.0281 M KMnO4, which oxidized the iron (II) to iron (III) while reducing the permanganate to manganese (II). Generate the balanced net ionic equation for the reaction. What is the mass percent of iron in this iron ore sample?
Answer:
a. MnO₄⁻ + 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O
b. 18.17% of Fe in the sample
Explanation:
a. In the reaction, Fe²⁺ is oxidized to Fe³⁺ and permanganate, MnO₄⁺ reduced to Mn²⁺, thus:
Fe²⁺ → Fe³⁺ + 1e⁻
MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺ + 4H₂O
5 times the iron and suming the manganese reaction:
MnO₄⁻ + 5e⁻ + 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + 5e⁻ + Mn²⁺ + 4H₂O
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂Ob. Moles of permanganate in the titration are:
0.03942L × (0.0281 moles / L) = 1.108x10⁻³ moles of MnO₄⁻
Based on the reaction, 1 mole of permanganate reacts with 5 moles of iron, if 1.108x10⁻³ moles of MnO₄⁻ reacts, moles of iron are:
1.108x10⁻³ moles of MnO₄⁻ × (5 moles Fe²⁺ / 1 mole MnO₄⁻) =
4.431x10⁻³ moles of Fe²⁺. Molar mass of Fe is 55.845g/mol. 4.431x10⁻³ moles of Fe²⁺ are:
4.431x10⁻³ moles of Fe²⁺ ₓ (55.845g / mol) =
0.2474g of Fe you have in your sample.Percent mass is:
0.2474g Fe / 1.362g sample ₓ 100 =
18.17% of Fe in the sampleThe mass percent of iron in the sample is 22.6%.
The net ionic equation of the reaction is;
5Fe^2+(aq) + 8H^+(aq) + MnO4^- -----> 5Fe^3+(aq) + Mn^2+(aq) + 4H2O(l)
Number of moles of MnO4^- = 39.42/1000 L × 0.0281 M = 0.0011 moles
If 5 moles of Fe^2+ reacts with 1 mole of MnO4^-
x moles of Fe^2+ reacts with 0.0011 moles
x = 5 moles × 0.0011 moles/1 mole
x = 0.0055 moles
Mass of Fe^2+ = 0.0055 moles × 56 g/mol = 0.308 g
Mass percent of iron = 0.308 g/ 1.362 g × 100/1
= 22.6%
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Provide two real world examples of habitats and niches within ecosystems? Explain the interdependence that exists within those ecosystems.
Answer:
An ecosystem encompasses living organisms and the nonliving elements of their environments. Hence, the components of an ecosystem include animals, plants, microorganisms, rocks, soil, minerals, atmosphere, and the surrounding water masses. An ecosystem can be huge, cutting across several nations, or it can be relatively small, such as the body of an animal, which is home to numerous microorganisms.
We can describe ecosystems under two headings – natural ecosystems and unnatural ecosystems. Unnatural ecosystems, which include agricultural and urban areas, are greatly modified and maintained by human activity. Conversely, natural ecosystems are self-sufficient, balanced ecological units, with a high proportion of native biodiversity and minimal human disruption.
The natural ecosystem is broad. It is divided into two major groups – terrestrial and aquatic ecosystems. These are further divided into many other smaller types of ecosystems as outlined in this article.
Explanation:
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What is the molecular mass of this substance? a. 31.02 u b. 63.02 u c. 47.02 u d. 126.04 u e. 110.01 u
Answer:
Molecular mass of HNO₃ = 63.015 g/mol
Explanation:
Note: The given question is incomplete , the given substance is nitric acid
Given:
Nitric acid (HNO₃)
Find:
Molecular mass
Computation:
Molecular mass of HNO₃ = 1.008 + 14.007 + 3(16)
Molecular mass of HNO₃ = 1.008 + 14.007 + 48
Molecular mass of HNO₃ = 63.015 g/mol
Which of the following is a characteristic of a scientific practice?
How would you expect cesium, Cs, to react with water? explaining your reasoning.
Answer:
Caesium reacts rapidly with water to form a colourless basic solution of caesium hydroxide (CsOH) and hydrogen gas (H2).
Explanation:
The reaction continues even when the solution becomes basic. The resulting solution is basic because of the dissolved hydroxide. The reaction is exothermic.
write anode and cathode in Zn-Ag galvanic cell
Explanation:
Zinc is the anode (solid zinc is oxidised). Silver is the cathode (silver ions are reduced).
By convention in standard cell notation, the anode is written on the left and the cathode is written on the right. So, in this cell: Zinc is the anode (solid zinc is oxidised). Silver is the cathode (silver ions are reduced).
what volume in liters of carbon monoxide will be required to produce 18.9 L of nitrogen in the reaction below
2co(g) + 2no(g) -> n2(g) + 2co2(g)
Answer:
37.8 L OF CARBON MONOXIDE IS REQUIRED TO PRODUCE 18.9 L OF NITROGEN.
Explanation:
Equation for the reaction:
2 CO + 2 NO ------> N2 + 2 CO2
2 moles of carbon monoxide reacts with 2 moles of NO to form 1 mole of nitrogen
At standard temperature and pressure, 1 mole of a gas contains 22.4 dm3 volume.
So therefore, we can say:
2 * 22.4 L of CO produces 22.4 L of N2
44.8 L of CO produces 22.4 L of N2
Since, 18.9 L of Nitrogen is produced, the volume of CO needed is:
44.8 L of CO = 22.4 L of N
x L = 18.9 L
x L = 18.9 * 44.8 / 22.4
x L = 18.9 * 2
x = 37.8 L
The volume of Carbon monoxide required to produce 18.9 L of N2 is 37.8 L
Answer:
37.8
Explanation:
Which of the following is the balanced reaction, given the rate relationships below.
a. rate = − 13 Δ[CH4] Δt = − 12 Δ[H2O] Δt = − Δ[CO2] Δt = 14 Δ[CH3OH] Δt
b. rate = − 12 Δ[N2O5] Δt = 12 Δ[N2] Δt = 15 Δ[O2] Δt
c. rate = − 12 Δ[H2] Δt = − 12 Δ[CO2] Δt = − Δ[O2] Δt = 12 Δ[H2CO3] Δt
Answer:
a. [tex]3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH[/tex]
b. [tex]2N_2O_5\rightarrow 2N_2 + 5O_2[/tex]
c. [tex]2H_2+2CO_2+O_2\rightarrow 2H_2CO_3[/tex]
Explanation:
Hello,
In this case, since those rate relationships have the stoichiometric coefficient at the denominators of the fractions preceding each ratio and the negative terms account for reactants and positive for products, we have:
a. [tex]3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH[/tex]
b. [tex]2N_2O_5\rightarrow 2N_2 + 5O_2[/tex]
c. [tex]2H_2+2CO_2+O_2\rightarrow 2H_2CO_3[/tex]
Best regards.
What is buffers and mention its importance?
Answer:
Buffer is the chemical substance that addition of acids and bases, maintaining constant environment,its called Buffer.
Explanation:
Buffers are use in the system to maintain the value of pH, and the contain the pH value is not to change.Buffer maintain the body of pH for the optimal activity,and they are solution of pH constant.Buffer in used in the lab and that to maintain growth of the micro tissues and the culture media.Buffer are used in maintain necessary optimal reaction activity,determine the indicator of solution with pH.Buffer capacity is that concentration to the buffering agent, is the very small increase,buffer capacity to the pH is 32% , of the maximum value of pH.Buffers in a acid regions to the desired of that value to the particular buffer agent.Buffers can be made from that a mixture of the base and acid, buffer can be a wide range of the obtained.Buffers that the pH calculation and they required to performed in the critic acid that the overlap over the buffer range.what is the bond energy required to break one mole of carbon-carbon bonds
Answer:
100 kcal of bond energy
in non- equilibrium cooling, carbon is allowed to diffuse its end position. true /false
Answer:
False because it is false
Predict the sign and calculate ΔS° for a reaction. Close Problem Consider the reaction H2CO(g) + O2(g)CO2(g) + H2O(l) Based upon the stoichiometry of the reaction the sign of Sºrxn should be _________ . Using standard thermodynamic data (in the Chemistry References), calculate Sºrxn at 25°C. Sºrxn = J/K•mol
Answer:
[tex]\mathbf{S^0_{rxn} = -140.41 \ J/mol.K}[/tex]
Based upon the stoichiometry of the reaction the sign of Sºrxn should be negative
Explanation:
Consider the reaction:
H2CO(g) + O2(g) --------> CO2(g) + H2O(l)
Using standard thermodynamic data;
Based upon the stoichiometry of the reaction the sign of Sºrxn should be _________ . calculate Sºrxn at 25°C. Sºrxn = J/K•mol
At standard thermodynamic data
[tex]\mathtt{S^0_{rxn} = \sum S^0 _{product} - \sum S^0 _{reactant}}[/tex]
[tex]S^0(CO_2)[/tex] = 213.79 J/mol.K
[tex]S^0(H_2O)=[/tex] 69.95 J/mol.K
[tex]S^0 ({H_2CO}) =[/tex] 218.95 J/mol.K
[tex]S^0 (O_2)[/tex] = 205.2 J/mol.K
[tex]\mathtt{S^0_{rxn} = (213.79 + 69.95) J/mol.K - (218.95+ 205.2) J/mol.K}[/tex]
[tex]\mathtt{S^0_{rxn} = (283.74) J/mol.K - (424.15) J/mol.K}[/tex]
[tex]\mathbf{S^0_{rxn} = -140.41 \ J/mol.K}[/tex]
Based upon the stoichiometry of the reaction the sign of Sºrxn should be negative
Al diluir 25 g de sal de mesa en 250ml de agua, ¿En cuántos °C aumenta el punto de ebullición de la disolución formada? ( Ke = 0,52 °C/molal , PM NaCl = 58,44 g/mol)
Answer:
ΔT=[tex]0.87^{\circ}C[/tex]
Explanation:
Para esta pregunta debemos recordar la ecuación que nos permite calcular el aumento ebulliscopico (aumento del punto de ebullición):
ΔT=[tex]Kb*m[/tex]
Donde ΔT es el valor del aumento del punto de ebullición. Kb es la constante ebulloscopica para el agua ([tex]0.512\frac{Kg~^{\circ}C}{mol}[/tex]) y m es la molalidad ([tex]m=\frac{mol}{Kg~ of~ solvente}[/tex]).
Por lo tanto el primer paso es calcular la molalidad de la solución. Para lo cual tendremos que calcular las moles de sal en los 25 g. Si queremos hacer esto debemos recordar que la formula de sal de mesa es NaCl y que la masa molar de NaCl es 58.44 g/mol. Por lo tanto:
[tex]25~g~NaCl\frac{1~mol~NaCl}{58.44~g~NaCl}~=~0.42~mol~NaCl[/tex]
Ahora bien, también debemos saber los Kg de agua en la solución. Por lo que podemos usar la densidad del agua (1 g/mL) para convertir de mL a g y luego hacer la conversión a Kg:
[tex]250~mL\frac{1~g}{1~mL}\frac{1~Kg}{1000~g}~=~0.25~Kg[/tex]
Finalmente para el calculo de la molalidad podemos dividir los dos valores:
[tex]m=\frac{0.42~mol}{0.25~Kg}=1.68[/tex]
Con el valor de la molalidad se puede calcular ΔT al reemplazar los valores:
ΔT=[tex]1.68~\frac{mol}{Kg}*0.52\frac{Kg~^{\circ}C}{mol}=0.87^{\circ}C[/tex]
Si la temperatura de ebullición normal del agua es 100 ºC. Podemos calcular la temperatura final si adicionamos ΔT:
Temperatura final = 100 + 0.87 = 100.87 ºC
Espero sea de ayuda!
Name the alkyne. Spelling and punctuation count.
The name of a compound is determined by IUPAC nomenclature
The International Union of Pure and Applied Chemistry have designated a universal rule for the nomenclature of organic compounds generally known as IUPAC nomenclature.
The whole idea of the IUPAC nomenclature is to have a universally agreed pattern of naming organic compounds according to their structure.
The compound is always named in such a way that the substituents and the functional group receives the lowest number.
In this case, the functional group is alkyne hence the name of the compound ends in -yne. The methyl group is substituted on position four.
Putting all these together the compound should be named as; 4-methylhept-2-yne
For IUPAC nomenclature of an inorganic compound
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A 1.0 L buffer solution is 0.250 M HC2H3O2 and 0.050 M LiC2H3O2. Which of the following actions will destroy the buffer?
A. adding 0.050 moles of NaOH
B. adding 0.050 moles of LiC2H3O2
C. adding 0.050 moles of HC2H3O2
D. adding 0.050 moles of HCl
E. None of the above will destroy the buffer.
Answer:
D
Explanation:
Addition of 0.05 M HCl, will react with all of the C2H3O2- from LiAc which will give 0.05 M more HAc. So there will be no Acetate ion left to make the solution buffer. Hence, the correct option for the this question is d, which is adding 0.050 moles of HCl.
The action that destroys the buffer is option c. adding 0.050 moles of HCl.
What is acid buffer?It is a solution of a weak acid and salt.
Here, The buffer will destroy at the time when either HC2H3O2 or NaC2H3O2 should not be present in the solution.
The addition of equal moles of HCl finishly reacts with equal moles of NaC2H3O2. Due to this, there will be only acid in the solution.
Since
moles of HC2H3O2 = 1*0.250 = 0.250
moles of NaC2H3O2 = 1*0.050 = 0.050.
moles of HCl is added = 0.050
Now
The reaction between HCl and NaC2H3O2
[tex]HCl + NaC_2H_3O_2 \rightarrow HC_2H_3O_2 + NaCl[/tex]
Now
BCA table is
NaC2H3O2 HCl HC2H3O2
Before 0.050 0.050 0.250
Change -0.050 -0.050 +0.050
After 0 0 0.300
Now, the solution contains the acid (HC2H3O2 ) only.
Therefore addition of 0.050 moles of HCl will destroy the buffer.
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An atom of 120In has a mass of 119.907890 amu. Calculate the mass defect (deficit) in amu/atom. Use the masses: mass of 1H atom
Answer:
a
Explanation:
answer is a on edg
Leo carefully pipets 50.0 mL of 0.500 M NaOH into a test tube. She places the test tube
into a small beaker to keep it from spilling and then pipets 75.0 mL of 0.250 M HCl into
another test tube. When Leo reaches to put this test tube of acid into the beaker along
with test tube of base she accidentally knocks the test tubes together hard enough to
break them and their respective contents combine in the bottom of the beaker. Is the
solution formed from the contents of the two test tubes acidic or basic? What is the pH of
the resulting solution?
Please answer below questions one by one to assist you receive full credits
(Alternatively, you can discard my hints below, solve the problem using your own way
and send me the picture/copy of your complete work through email)
The mole of NaOH before mixing is
mol (save 3 significant figures)
The mole of HCl before mixing is
mol (save 4 significant figures)
After mixing, the solution is
(choose from acidic or basic)
The total volume of mixture is
L (save 3 significant figures)
The concentration of [OH-] is
M (save 3 significant figures)
The concentration of [H'l is
M (save 3 significant figures)
Let's consider the neutralization reaction between HCl and NaOH.
NaOH + HCl ⇒ NaCl + H₂O
To determine the pH of the resulting mixture, we need to determine the reactant in excess. First, we will calculate the reacting moles of each reactant.
NaOH: 0.0500 L × 0.500 mol/L = 0.0200 mol
HCl: 0.0750 L × 0.250 mol/L = 0.0188 mol
Now, let's determine the reactant in excess and the remaining moles of that reactant.
NaOH + HCl ⇒ NaCl + H₂O
Initial 0.0200 0.0188
Reaction -0.0188 -0.0188
Final 1.20 × 10⁻³ 0
The volume of the mixture is 50.0 mL + 75.0 mL = 125.0 mL. Then, 1.20 × 10⁻³ moles of NaOH are in 125.0 mL of solution. The concentration of NaOH is:
[NaOH] = 1.20 × 10⁻³ mol/0.1250 L = 9.60 × 10⁻³ M
NaOH is a strong base according to the following equation.
NaOH ⇒ Na⁺ + OH⁻
The concentration of OH⁻ is 1/1 × 9.60 × 10⁻³ M = 9.60 × 10⁻³ M.
The pOH is:
pOH = -log [OH⁻] = -log 9.60 × 10⁻³ = 2.02
We will calculate the pH using the following expression.
pH = 14.00 - pOH = 14.00 - 2.02 = 11.98
The pH is 11.98. Since pH > 7, the solution is basic.
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According to the following reaction, how many grams of ammonia will be formed upon the complete reaction of 31.2 grams of hydrogen gas with excess nitrogen gas ? nitrogen(g) + hydrogen(g) ammonia(g)
Answer:
176.8 g of ammonia, NH3.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
N2 + 3H2 —> 2NH3
Next, we shall determine the mass of H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:
Molar mass of H2 = 2x1 = 2 g/mol
Mass of H2 from the balanced equation = 3 x 2 = 6 g
Molar mass of NH3 = 14 + (3x1) = 17 g/mol
Mass of NH3 from the balanced equation = 2 x 17 = 34 g.
From the balanced equation above,
6 g of H2 reacted to produce 34 g of NH3.
Finally, we shall determine the mass of ammonia, NH3 produced by reacting 31.2 g of H2.
This can be obtained as follow:
From the balanced equation above,
6 g of H2 reacted to produce 34 g of NH3.
Therefore, 31.2 g of H2 will react to produce = (31.2 x 34)/6 = 176.8 g of NH3.
Therefore, 176.8 g of ammonia, NH3 were obtained from the reaction.