# How do you apply gas stoichiometry?

Mar 23, 2014

${N}_{2}$ + 3${H}_{2}$ ---> 2N${H}_{3}$ What volume of hydrogen is necessary to react with five liters of nitrogen to produce ammonia? (Assume constant temperature and pressure.)

${N}_{2}$ + 3${H}_{2}$ ---> 2N${H}_{3}$ , as per this equation, 1 volume of Nitrogen , needs 3 volumes of Hydrogen. The reaction produces two volumes of ammonia.

Let us write this as equation;

$\text{1 volume of Nitrogen"/"3 volume of Hydrogen}$ (a)

If we start with 5 L of Nitrogen , X volumes of Hydrogen will be needed.

Let us write this as equation;

$\text{5 L of Nitrogen"/"X L of Hydrogen}$ (b)

equate (a) and (b)
$\frac{1}{3}$ = $\frac{5}{X}$

cross multiply $1 \cdot X$ = $3 \cdot 5$ , X=15 L

2CO + ${O}_{2}$ ---> 2C${O}_{2}$ How many liters of carbon dioxide are produced if 75 liters of carbon monoxide are burned in oxygen? How many liters of oxygen are necessary?

As per this equation, 2 volumes of Carbon Monoxide, needs 1 volume of Oxygen. The reaction produces two volumes of Carbon dioxide.

Let us write this as equation;

$\text{2 volume of Carbon Monoxide"/"2 volumes of Carbon Dioxide}$ (a)

If we start with 75 L of Carbon monoxide , X volumes of carbon dioxide will be produced.

Let us write this as equation;

"75 L of CO"/("X L of "CO_2) (b)

equate (a) and (b)

$\frac{2}{2}$ = $\frac{75}{X}$

cross multiply $2 \cdot X$ = $2 \cdot 75$ , X= 75 L

$\text{2 volume of Carbon Monoxide"/"1 volumes of Oxygen}$ (a)

If we start with 75 L of Carbon Monoxide , X volumes of Oxygen will be needed.

"75 L of CO"/("X L of "O_2) (b)

equate (a) and (b)

$\frac{2}{1}$= $\frac{75}{X}$

$2 \cdot X$ = $75 \cdot 1$

2X =75

X= 37.5 L