Question

2 H2(g) + O2(g) → 2 H2O(g)If 40 L of hydrogen reacts with 20 L of oxygen, then the volume of water (in L) produced is

Answer 1

2 H2(g) + O2(g) → 2 H2O(g) (1)

Don't forget to balance your reaction.

First, we need to determine which is the limiting reactant, H2 or O2.

Before we are doing this, we must assume STP conditions: temperature of 273 K (0° Celsius or 32° Fahrenheit) and the standard pressure of 1 atm.

Under these conditions, one mole of a gas occupies 22.4 L.

The Limiting Reactant:

Use stoichiometry and (1)

Remember: STP => 1 mol of gas = 22.4 L

2 x 22.4 L of H2 ----------- 22.4 L of O2

40 L of H2 ----------- x

`x\text{ = }\frac{40L\text{ x 22.4 L}}{2x22.4\text{ L}}=20\text{ L}`

From this we say the limiting reactant is O2, we have exactly 20 L of this gas, and stoichiometry says that we need 20L.

The volume of water:

We use O2 to determine the volume of water,

22.4 L of O2 --------- 2 x 22.4 L of water

20 L of O2 --------- y

y = 40 L

Answer: 40L of water