How many liters of oxygen are required to react completely with 1.2 liters of hydrogen to form water?

Jun 14, 2016

Under the same conditions a volume of $0.6 \cdot L$ of dioxygen is required.

Explanation:

${H}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow {H}_{2} O \left(l\right)$

Under equivalent conditions, equal volumes of gases contain equal numbers of gaseous particles. We need 1 equiv ${O}_{2} \left(g\right)$ for each 2 equiv ${H}_{2} \left(g\right)$.

Jun 14, 2016

$0.6 L {O}_{2}$

Explanation:

Without the pressure and temperature we cannot calculate this value.

However, if we assume this reaction is taking place at Standard Temperature and Pressure (STP) we can use Avogadro's constant of $22.4 L$ per mole.

The balanced equation for this reaction is

$2 {H}_{2} + {O}_{2} \to 2 {H}_{2} O$

Now we can use the mole ratio to convert the values.

$1.2 L {H}_{2} \cdot \frac{1 m o l {H}_{2}}{22.4 L {H}_{2}} \cdot \frac{1 m o l {O}_{2}}{2 m o l {H}_{2}} \cdot \frac{22.4 L {O}_{2}}{1 m o l {O}_{2}}$

$1.2 \cancel{L {H}_{2}} \cdot \frac{1 \cancel{m o l {H}_{2}}}{\cancel{22.4} \cancel{L {H}_{2}}} \cdot \frac{1 \cancel{m o l {O}_{2}}}{2 \cancel{m o l {H}_{2}}} \cdot \frac{\cancel{22.4} L {O}_{2}}{1 \cancel{m o l {O}_{2}}}$

$0.6 L {O}_{2}$