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If 0.00516 mols of HC2H302 are reacted with 0.00585 mols of NaHCO3, which is the limiting reactant?
Show your work.
NaHCO3(s) + HC2H302(aq) → C2H302Na(aq) +H2CO3(aq) H2CO3(aq) - H2O(1) +CO2(g)

Answers

Answer 1

Look at the reaction

More than 1mol of HC2H3O2 is required so the mols will be doubled atleast

0.00516(2)=0.01032mol

Now

NaHCO_3 is less

So Sodium bicarbonate is the limiting reagent
Answer 2

In the given query, the limiting reagent in provided chemical reaction is: [tex]\rm NaHCO_3(s) + HC_2H_3O_2(aq) \rightarrow C_2H_3O_2Na(aq) +H_2CO_3(aq)+ H_2O(l) +CO_2(g)[/tex] is [tex]\rm HC_2H_3O_2[/tex].

In a reaction, limiting reagent is a chemical substance that is completed used or is totally consumed resulting into limiting the formation of further products.

In the provided reaction:

[tex]\rm NaHCO_3(s) + HC_2H_3O_2(aq) \rightarrow C_2H_3O_2Na(aq) +H_2CO_3(aq)+ H_2O(l) +CO_2(g)[/tex]

Given: 0.00516 mol [tex]\rm HC_2H_3O_2[/tex] and 0.00585 mol [tex]\rm NaHCO_3[/tex].

Using balanced equation, the stoichiometric ratio of  [tex]\rm HC_2H_3O_2[/tex] to  [tex]\rm NaHCO_3[/tex] is 1:1. Meaning that, for every 1 mole of  [tex]\rm HC_2H_3O_2[/tex], we need 1 mole of  [tex]\rm NaHCO_3[/tex].

Therefore, the number of moles of  [tex]\rm NaHCO_3[/tex] needed to react with 0.00516 mol of  [tex]\rm HC_2H_3O_2[/tex] is:

0.00516 mol  [tex]\rm HC_2H_3O_2[/tex] × (1 mol  [tex]\rm NaHCO_3[/tex]/ 1 mol  [tex]\rm HC_2H_3O_2[/tex])

= 0.00516 mol [tex]\rm NaHCO_3[/tex].

Since we have 0.00585 mol of [tex]\rm NaHCO_3[/tex], which is greater than the amount needed (0.00516 mol), [tex]\rm NaHCO_3[/tex] is not the limiting reactant.

Calculate the number of moles of [tex]\rm HC_2H_3O_2[/tex]needed to react with 0.00585 mol of [tex]\rm NaHCO_3[/tex]:

0.00585 mol [tex]\rm NaHCO_3[/tex] × (1 mol [tex]\rm HC_2H_3O_2[/tex]/ 1 mol  [tex]\rm NaHCO_3[/tex]) = 0.00585 mol  [tex]\rm HC_2H_3O_2[/tex]

As we see, only 0.00516 mol of  [tex]\rm HC_2H_3O_2[/tex] are present, which is less than the amount needed (0.00585 mol),  [tex]\rm HC_2H_3O_2[/tex] is the limiting reactant.

Therefore,  [tex]\rm HC_2H_3O_2[/tex]is the limiting reactant in the provided chemical reaction.

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Answers

Answer:

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Answers

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Stoichiometric calculation

Looking at the equation of the reaction:

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The mole ratio of CuBr2 and NaCl is 1:2.

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Answers

Answer:

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Explanation:

the only reason i can think of why it wouldn't be 11.111 is improper rounding and sig figs. so if you round with sig figs it would be 11.1

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Answers

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Answers

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Answers

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Answers

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Answers

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Limiting reactants

From the equation of the reaction:

[tex]MnF_4(aq)+2(NH_4)_2S(aq)- > MnS_2(s)+4NH_4F(aq)[/tex]

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