Q. 7.66 - An Introduction to General, Organic, and Biological Chemistry

Question

When the gases dihydrogen sulfide and oxygen react, they form the gases sulfur dioxide and water vapor.

a. Write the balanced chemical equation for the reaction.

b. How many grams of oxygen are needed to react with $2.50 \mathrm{g}$ of dihydrogen sulfide?

c. How many grams of sulfur dioxide can be produced when $38.5 \mathrm{g}$ of oxygen reacts?

d. How many grams of oxygen are needed to produce$55.8 \mathrm{g}$ of water vapor?

Step-by-Step Solution

Answer

(a) The balanced chemical equation of this equation is $2H_{2}S\left(g\right)+3O_2\left(g\right)\to 2S{O}_2\left(g\right)+2H_{2}O\left(g\right)$.

(b) The $2.35 \mathrm{g}$of oxygen are required to react with $2.50 \mathrm{g}$dihydrogen sulfide.

(c) The $50.7 \mathrm{g}$of sulfur dioxide can produced when $38.5 \mathrm{g}$of oxygen reacts.

(d) The $1.54 \mathrm{mol} {\mathrm{O}}_2\left(\mathrm{g}\right)$ of oxygen are needed to produce $55.8 \mathrm{g}$of water vapor.

Part(a) Step 1: Given information

We have been given,

The gases dihydrogen sulfide and oxygen react, they form the gases sulfur dioxide and water vapor. 

Part(a) Step 2: Explanation

We know that the balanced chemical equation for the reaction is :

When the gases dihydrogen sulfide and oxygen reacts, they form the gases sulfur dioxide

and water vapor.

$2H_{2}S\left(g\right)+3O_2\left(g\right)\to 2S{O}_2\left(g\right)+2H_{2}O\left(g\right)$

Part(b) Step 1: Given information

We have to find out how many grams of oxygen are needed to react with $2.50 \mathrm{g}$of dihydrogen sulfide.

Part(b) Step 2: Explanation

Now, we calculate grams of oxygen are needed to react with $2.50 \mathrm{g}$dihydrogen sulfide,

${\mathrm{H}}_2\mathrm{S}+{\mathrm{O}}_2\to {\mathrm{SO}}_2+{\mathrm{H}}_2 \mathrm{moles} {\mathrm{H}}_2\mathrm{S}=\frac{2.50 \cancel{\mathrm{g}}}{34.082 \cancel{\mathrm{g}}/\mathrm{mol}}=0.0734 \mathrm{mol}$

$\mathrm{moles} {\mathrm{O}}_2 \mathrm{mass} {\mathrm{O}}_2=0.0734 \cancel{\mathrm{mol}}\times \raisebox{1ex}{$32\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\cancel{\mathrm{mol}}$}\right.=2.35\mathrm{g}$

Hence $2.35 \mathrm{g}$of oxygen required to react with $2.50 \mathrm{g}$dihydrogen sulfide.

Part(c) Step 1: Given information

We have to find out the how many grams of sulfur dioxide can be produced when $38.5 \mathrm{g}$of oxygen reacts. 

Part(c) Step 2: Explanation

When the gases dihydrogen sulfide and oxygen react, they form the gases sulfur dioxide and water vapor.

$2H_{2}S\left(g\right)+3O_2\left(g\right)\to 2S{O}_2\left(g\right)+2H_{2}O\left(g\right)$

$[38 \cancel{\mathrm{g}} \cancel{{\mathrm{O}}_2}]\times \frac{[1 \cancel{\mathrm{mol}} \cancel{{\mathrm{O}}_2}]}{[32 \cancel{\mathrm{g}} \cancel{{\mathrm{O}}_2}]}\times \frac{[2 \cancel{\mathrm{mol}} \cancel{{\mathrm{SO}}_2}]}{[3 \cancel{\mathrm{mol}} \cancel{{\mathrm{O}}_2}]}\times \frac{[64 {\mathrm{gSO}}_2]}{[1 \cancel{\mathrm{mol}} \cancel{{\mathrm{SO}}_2}]}=50.7 \mathrm{g} {\mathrm{SO}}_2$

Hence $50.7 \mathrm{g} {\mathrm{SO}}_2$can be produced when $38.5 \mathrm{g}$ of oxygen reacts.

Part(d) Step 1: Given information

We have to find out the how many grams of oxygen are needed to produce $55.8 \mathrm{g}$of water vapor.

Part(d) Step 2: Explanation

Now, we calculate how many grams of oxygen are required to produce $55.8 \mathrm{g}$of water vapor,

$H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to H_{2}O\left(g\right)$,

Moles of water$=\frac{55.8 \mathrm{g}}{18.01 \mathrm{g}/\mathrm{mol}}=3.09 \mathrm{mol}$,

If there are $3.09 \mathrm{mol}$of water, then clearly there were $1.54 \mathrm{g}/\mathrm{mol} {\mathrm{O}}_2\left(\mathrm{g}\right)$as a reactant,

And thus mass of oxygen $=1.54 \cancel{\mathrm{mol}}\times 32.00 \frac{\mathrm{g}}{\cancel{\mathrm{mol}}}=55.8 \mathrm{g}$