Chapter 7, Problem 7.81QA

Interpretation Introduction

To find:

a. Limiting reactant in a reaction mixture of $775 g CH{Cl}_3$ and $775 g HF$(to solve this problem, we need to consider $775 g of HF$ instead of $775 g of HCl$), where the reactions are:

(i) $CH{Cl}_3\left(l\right)+2 HF\left(g\right)\to {CHClF}_2\left(g\right)+2 HCl\left(g\right)$

(ii) $2 {CHClF}_2\left(g\right)\to C_2{F}_4\left(g\right)+2 HCl\left(g\right)$.

b. How many grams of $C_2{F}_4$ could be prepared in the reaction (ii) from the product formed in reaction (i), where the yield in reaction (i) is stoichiometric and in reaction (ii) is $95\%$.

c. How much of which reactant is left over in the reaction mixture in part (i)?

Answer to Problem 7.81QA

Solution:

a. Limiting reactant in a reaction mixture of $775 g CH{Cl}_3$ and $775 g HF$ is $CH{Cl}_3$.

b. ${308 g C}_2{F}_4$ could be prepared in the reaction (ii) from the product formed in reaction (i), where the yield in reaction (i) is stoichiometric and in reaction (ii) is $95\%$.

c. $515 g HF$ is left over in the reaction mixture in part (i).

Explanation of Solution

1) Concept:

Limiting reactant: It is defined as the reactant, which is completely consumed in the chemical reaction and defines the amount of product produced.

Excess of reactant: The reactant, which is left over to some extent in unreacted form after completion of reaction.

Amount of excess of reactant can be calculated by subtracting the amount reacted from the initial amount of reactant.

i.e., $Unreacted mass=Initial mass-reacted mass$

Percent yield is the percent of amount of product formed.

i.e.,

$Percent yield=\frac{Actual yield}{Theoretical yield} \times 100$

where,

$Actual yield$= Amount of product obtained from the actual reaction (experiment).

$Theoretical yield=$ Amount of product obtained from stoichiometric calculation (i.e., from the limiting reactant).

2) Formula:

$Unreacted mass=Initial mass-reacted mass$

$Percent yield=\frac{Actual yield}{Theoretical yield} \times 100$

3) Given:

i) $775 g CH{Cl}_3$

ii) $775 g HF$(to solve this problem, we need to consider $775 g of HF$ instead of $775 g of HCl.$)

iii) $Percent yield for the second reaction =95\%$

4) Calculation:

a. Limiting reactant:

Given reactions are:

i) $CH{Cl}_3\left(l\right)+2 HF\left(g\right)\to {CHClF}_2\left(g\right)+2 HCl\left(g\right)$

ii) $2 {CHClF}_2\left(g\right)\to C_2{F}_4\left(g\right)+2 HCl\left(g\right)$.

Reaction mixture contains $775 g CH{Cl}_3$ and $775 g HF$(to solve this problem, we need to consider $775 g of HF$ instead of $775 g of HCl.$)

Calculate the moles of both $CH{Cl}_3$ and $HF$(to solve this problem, we need to consider $775 g of HF$ instead of $775 g of HCl$)

$775 g CH{Cl}_3\times \frac{1 mol }{119.38 g}=6.492 mol CH{Cl}_3$

$775 g HF\times \frac{1 mol }{20.01 g}=38.73 mol HF$

(To solve this problem, we need to consider $775 g of HF$ instead of  $775 g of HCl.$)

Calculate the mass of $CHCl{F}_2$ produced from both the reactants:

The mole ratio between $CHC{l}_3$ and $CHCl{F}_2$ is $1:1$

$6.492 mol CH{Cl}_3\times \frac{1 mol {CHClF}_2 }{1 mol CH{Cl}_3}\times \frac{86.47 g}{1 mol}= 561.36 g CHCl{F}_2$

The mole ratio between $CHC{l}_3$ and $HF$ is $1:2$

 (To solve this problem, we need to consider $775 g of HF$ instead of  $775 g of HCl.$)

$38.73 mol HF\times \frac{1 mol {CHClF}_2}{2 mol HF}\times \frac{86.47 g}{1 mol}=1674.49 g CHCl{F}_2$

(To solve this problem, we need to consider $775 g of HF$ instead of  $775 g of HCl$.)

The amount of $CHCl{F}_2$ produced by $CH{Cl}_3$ is less; hence, the limiting reactant in this reaction mixture is $CH{Cl}_3$.

b. Mass of $C_2{F}_4$:

From part a, we know that $6.492 mol$${CHClF}_2$ is produced from $CH{Cl}_3$.

From reaction (ii), the mole ratio of ${CHClF}_2:C_2{F}_4$ is 2:1, thus, calculating grams of $C_2{F}_4$:

$6.492 mol CHCl{F}_2\times \frac{1 mol C_2{F}_4}{2 mol CHCl{F}_2}\times \frac{100.014 g{ C}_2{F}_4 }{1 mol C_2{F}_4}=324.64 g C_2{F}_4$

This is the theoretical yield, but the yield from (ii) is 95%, so the formula for percent yield is

$Percent yield=\frac{Actual yield}{Theoretical yield} \times 100$

Re-arranging it for actual yield,

$Actual yield=\frac{percent yield*theoretical yield}{100}$

$=\frac{95*324.64 g}{100}$

$=308 g$

c. Mass of excess of reactant in reaction (i):

(To solve this problem, we need to consider $775 g of HF$ instead of $775 g of HCl.$)

Calculate the mass of $HF$ required for reaction as

Mole ratio of $CH{Cl}_3:HF$ is 1:2

$6.492 mol CH{Cl}_3\times \frac{2 mol HF }{1 mol CH{Cl}_3}\times \frac{20.01 g}{1 mol}=259.81g HF$

(To solve this problem, we need to consider $775 g of HF$ instead of $775 g of HCl.$)

Calculate the amount of $HF$ remaining unreacted:

$Unreacted mass=Initial mass-reacted mass$

$Unreacted mass of HF=775 g-259.81g$

$=515.19 g HF$

$=515 g HF$(in correct significant figures)

Therefore, $515 g HF$ is left over in the reaction mixture in part (i)

(To solve this problem, we need to consider $775 g of HF$ instead of $775 g of HCl.$)

Conclusion:

We determined the limiting reactant from the reactant’s masses and found out the excess of reactant. Using the percent yield, actual yield is calculated.