Question

Assuming all volume measurements are made at the same temperature and pressure, how many liters of water vapor can be produced when 8.23 liters of oxygen gas react with excess hydrogen gas? Show all of the work used to solve this problem.

Unbalanced equation: H2 (g) + O2 (g) H2 O(g)

Answer 1

the reaction of oxygen gas with excess hydrogren would be:

2H2 (g) + O2 (g) --> 2H2O (g)
1 mol of O2 yields 2 mol H2O.
Likewise, 1 L of O2 yields 2L H2O.
so, 8.23 L O2 yields 16.46 L H20

hope this help

Answer 2

Answer : The volume of water vapor is, 16.46 liters

Explanation :

The balanced chemical reaction will be,

$2H_2(g)+O_2(g)\rightarrow 2H_2O(g)$

The mole ratio of hydrogen, oxygen and water vapor are 2 : 1 : 2.

According to the Avogadro's Law, the volume is directly proportional to the number of moles of the gas at constant pressure and temperature.

$V\propto n$   (At constant temperature and pressure)

or,

$\frac{V_1}{V_2}=\frac{n_1}{n_1}$

where,

$V_1$ = volume of oxygen = 8.23 L

$V_2$ = volume of water vapor = ?

$n_1$ = moles of oxygen = 1 mole

$n_2$ = moles of water vapor = 2 mole

Now put all the given values in the above formula, we get the volume of water vapor.

$\frac{8.23L}{V_2}=\frac{1mole}{2mole}$

$V_2=16.46L$

Therefore, the volume of water vapor is, 16.46 liters