Chapter 10, Problem 10.128QA

Interpretation Introduction

To find:

The total pressure of the gases released by the reaction of $H_2{O}_2$ and $N_2{H}_4$ by using the reaction stoichiometry.

Answer to Problem 10.128QA

Solution:

Total pressure of the gases released by the reaction is $0.567 atm$.

Explanation of Solution

1) Concept:

To calculate the total pressure exerted by the gases released in the reaction, we can use gaseous stoichiometry. We calculate the moles of the $H_2{O}_2$  and $N_2{H}_4$ from the mass of the liquid and the molar masses. From the moles of the reactant, we calculate the limiting reactant by using the reaction stoichiometry. We will calculate the moles of the gases product from the limiting reactant and then calculate the total pressure by using the ideal gas equation.

2) Formula:

i) $PV=nRT$      

ii) $Molar mass=\frac{mass}{mol}$

iii) $P_{total}= P_{H_2{O}_2}+ P_{N_2{H}_4}$

3) Given:

i) $Mass of H_2{O}_2 = 8.50 g$

ii) $Mass of N_2{H}_4 = 4.00 g$

iii) Volume of the reaction vessel $V=125 L$

iv)$T=1110 ℃=1110+273.15=1383.15 K$

v) $R=0.08206 \frac{L. atm}{mol. K}$

4) Calculations:

i) The balanced chemical reaction is

$2{ H}_2{O}_2 \left(l\right) +N_2{H}_4 \left(l\right) \to N_2\left(g\right)+4 { H}_{2}O \left(g\right)$

Calculating moles of the ${ H}_2{O}_2$ and $N_2{H}_4$ using their molar masses as

$8.50 g { H}_2{O}_2 \times \frac{1 mol}{34.01 g { H}_2{O}_2}=0.2499 mol { H}_2{O}_2$

$4.00 g N_2{H}_4 \times \frac{1 mol}{32.04 g N_2{H}_4 }=0.1248 mol N_2{H}_4$

Calculatingthe limiting reactant as

$0.2499 mol { H}_2{O}_2 \times \frac{1 mol N_2 }{2 mol { H}_2{O}_2}=0.12495 mol N_2$

$0.1248 mol N_2{H}_4 \times \frac{1 mol N_2 }{1 mol N_2{H}_4}=0.1248 mol N_2$

So, the limiting reactant is $N_2{H}_4$.

ii) Calculatingthe moles of the gases product from the moles of the $N_2{H}_4$

$0.1248 mol N_2{H}_4 \times \frac{1 mol N_2 }{1 mol N_2{H}_4}=0.1248 mol N_2$

$0.1248 mol N_2{H}_4 \times \frac{4 mol H_{2}O }{1 mol N_2{H}_4}=0.4992 mol H_{2}O$

iii) Calculatingthe partial pressure of ${ H}_2{O}_2$ and $N_2{H}_4$ by using ideal gas equation

$PV=nRT$

$P_{N_2}=\frac{nRT}{V}= \frac{0.1248 mol\times 0.08206 \frac{L. atm}{mol. K}\times 1383.15 K }{125 L}=0.1133 atm$

$P_{H_{2}O}=\frac{nRT}{V}= \frac{0.4992 mol\times 0.08206 \frac{L. atm}{mol. K}\times 1383.15 K }{125 L}=0.4533 atm$

$P_t= P_{H_2{O}_2}+ P_{N_2{H}_4}=0.1133 atm+0.4533 atm=0.567 atm$

The total pressure of the gases released by the reaction is $0.567 atm$.

Conclusion:

The total pressure of the gas released from the reaction is calculated from the mass of the reactant and the reaction stoichiometry by using the ideal gas equation to calculate the partial pressure.