Chapter 6, Problem 6.109P

Interpretation Introduction

(a)

Interpretation:

For the given reaction, the limiting reactant should be determined.

Concept Introduction:

Molarity of solution is defined as number of moles of solute in 1 L of solution.

It is mathematically represented as follows:

$M=\frac{n}{V}$

Here, n is number of moles and v is volume of solution in L.

In a chemical reaction, limiting reactant is one which is totally consumed in the completion of the reaction. It limits the amount of product formed in the reaction.

Answer to Problem 6.109P

Hydrogen bromide, HBr is the limiting reactant.

Explanation of Solution

The equation should be balanced first,

$\text{Ca(s)+HBr(aq)}\to {\text{CaBr}}_{\text{2}}{\text{(aq)+H}}_{\text{2}}\text{(g)}$

To balance the number of hydrogen and bromine give coefficient 2 to HBr thus,

$\text{Ca(s)+ 2HBr(aq)}\to {\text{CaBr}}_{\text{2}}{\text{(aq)+H}}_{\text{2}}\text{(g)}$

To determine the limiting reactant, calculate the number of moles of Ca and HBr present.

The molarity of HBr is 0.325 m and volume is 115 mL. Now, number of moles of HBr can be calculated as follows:

$M=\frac{n}{V}$

Rearranging,

$n=M\times V$

Putting the values,

$\begin{array}{c}{n}_{\text{H}Br}=\left(0.325\text{ M}\right)\left(\frac{1\text{ mol/L}}{1\text{ M}}\right)\left(115\text{ mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\\ =0.03737\text{ mol}\end{array}$

Similarly, number of moles of Ca can be calculated as follows:

$n=\frac{m}{M}$

Molar mass of Ca is 40.078 g/mol thus,

$\begin{array}{c}{n}_{\text{C}\text{a}}=\frac{1.46\text{ g}}{40.078\text{ g/mol}}\\ =0.0364\text{ mol}\end{array}$

From the balanced chemical reaction,

$\text{Ca(s)+ 2HBr(aq)}\to {\text{CaBr}}_{\text{2}}{\text{(aq)+H}}_{\text{2}}\text{(g)}$

1 mol Ca reacts with 2 mol of HBr thus, for 0.0364 mol of Ca, 0.07285 mol of HBr is required. But number of moles of HBr is 0.03737 mol thus, HBr is a limiting reactant.

Interpretation Introduction

(b)

Interpretation:

The volume of hydrogen gas produced should be calculated if the vapor pressure of water at $22{}^{\circ }\text{C}$ with a total pressure of 754 torr.

Concept Introduction:

The ideal gas equation is as follows:

$PV=nRT$

Here, p is pressure, v is volume, n is number of moles, r is Universal gas constant and t is temperature.

Answer to Problem 6.109P

0.4560 L.

Explanation of Solution

The balanced chemical reaction is as follows:

$\text{Ca(s)+ 2HBr(aq)}\to {\text{CaBr}}_{\text{2}}{\text{(aq)+H}}_{\text{2}}\text{(g)}$

Since, HBr is a limiting reactant thus, amount of hydrogen gas produced depends on the amount of HBr.

From the balanced chemical reaction, 2 mol of HBr gives 1 mol of ${\text{H}}_{\text{2}}$

Thus, 1 mol of HBr gives 0.5 mol of ${\text{H}}_{\text{2}}$

The number of HBr is 0.03737 mol thus, number of moles of ${\text{H}}_{\text{2}}$ will be 0.018685 mol.

From ideal gas equation, volume can be calculated as follows:

$V=\frac{nRT}{P}$

Pressure is 754 torr or 0.9921 atm and temperature is $22{}^{\text{O}}\text{C}$ or 295.15 K.

Putting the values,

$\begin{array}{c}V=\frac{\left(0.018695\text{ mol}\right)\left(0.082{\text{ L atm mol}}^{-1}{\text{ K}}^{-1}\right)\left(295.15\text{ K}\right)}{\left(0.9921\text{ atm}\right)}\\ =0.4560\text{ L}\end{array}$

Thus, volume of hydrogen gas is 0.4560 L.

Interpretation Introduction

(c)

Interpretation:

The mass of excess reactant remain after the completion of reaction should be calculated.

Concept Introduction:

The excess reactant is the one that present in excess amounts, the amount of product does not depend on the amount of the excess reactant and it remains after the completion of the reaction.

Answer to Problem 6.109P

$\text{0}\text{.71018 g}$ of Ca remains after the completion of reaction.

Explanation of Solution

The balanced chemical reaction is as follows:

$\text{Ca(s)+ 2HBr(aq)}\to {\text{CaBr}}_{\text{2}}{\text{(aq)+H}}_{\text{2}}\text{(g)}$

Here, HBr is limiting reactant, the number of moles of HBr is 0.03737 mol. From the reaction, 2 mol of HBr reacts with 1 mol of Ca thus, 1 mol of HBr reacts with 0.5 mol of Ca. Thus, 0.03737 mol HBr reacts with 0.018685 mol of Ca. The number of moles of Ca present is 0.0364 mol thus, number of moles of Ca remains can be calculated as follows:

$\begin{array}{c}{n}_{\text{C}\text{a},r\text{e}m\text{a}ining}=0.0364-0.018685\\ =0.01772\text{ mol}\end{array}$

Since, molar mass of Ca is 40.078 g.mol, its mass can be calculated as follows:

$m=n\times M$

Putting the values,

$\begin{array}{c}m=0.01772\text{ mol}\times 40.078\text{ g/mol}\\ \text{=0}\text{.71018 g}\end{array}$.