# Question fd292

Oct 29, 2016

Balanced equation of the given reaction

$2 M o {S}_{2} + 7 {O}_{2} \left(g\right) \to 2 M o {O}_{3} \left(s\right) + 4 S {O}_{2} \left(g\right)$

Atomic masses considered

$M o \to 96 \text{ g/mol}$

$S - 32 \text{ g/mol}$

$O \to 16 \text{ g/mol}$

Molar masses calculated

${O}_{2} \to 2 \times 16 = 32 \text{ g/mol}$

$M o {O}_{3} \to 96 + 3 \times 16 = 144 \text{ g/mol}$

Mass of $M o {O}_{3}$ to be produced

$= 1.82 k g = 1820 g = \frac{1820 g}{144 \frac{g}{\text{mol}}} = 12.64 m o l$

By the stochiometric ratio of reactants and products in the balanced equation of the reaction involved we can say

To produce 2 moles of $M o {O}_{3}$ we require 7 moles ${O}_{2} \left(g\right)$
So
To produce 12.64 moles of $M o {O}_{3}$ we require

$= \frac{7}{2} \times 12.64 = 18.96 \text{ moles} {O}_{2} \left(g\right)$

Given

$P \to \text{Pressure of "O_2(g)=715"torr} = \frac{715}{760} a t m$

$T \to \text{Temperature of } {O}_{2} \left(g\right) = {38}^{\circ} C = 273 + 38 = 311 K$

$n \to \text{No. of moles of "O_2(g)=18.96 " moles}$

$R \to \text{Universal gas constant} = 0.082 L a t m {K}^{-} 1 m o {l}^{-} 1$

V-> "Volume of "O_2(g)=?#

Inserting these values in the equation of state for ideal gas we get

$P V = n R T$

$V = \frac{n R T}{P} = \frac{18.96 m o l \times 0.082 L a t m {K}^{-} 1 m o {l}^{-} 1 \times 311 K}{\frac{715}{760} a t m}$

$= 513.95 L$