Question

If 2 moles of water decomposes, the products that will form are 2 moles of hydrogen gas and one mole of oxygen gas. If a reaction is conducted that consumes 20.0 mL of water, what volume of oxygen gas will be produced

Answer 1

Step-by-step explanation:

The balanced equation for the decomposition of water is:

2H2O = 2H2 + 1O2

This tells us that we should expect 1 mole of O2 for every 2 moles of H2O, a molar ratio of (1 O2/2 H2O).

If we start with 2 moles of water, we should get (2mole H2O)*(1 O2/2 H2O) = 1 mole of H2.

But we are given just 20ml of water by the stingy boss. We need to calculate the number of moles of water in that 20 ml. We'll assume pure water at room temperature, so density = 1g/ml. We have 20 g of H2O. To find moles, divide by water's molar mass of 18 g/mole:

20g/(18g/mole) = 1.11 moles H2O

The moles of H2 produced would be:

(1.11 moles H2O)*(1 O2/2 H2O) = 0.55 moles O2 produced

We are not given the conditions of temperature and pressure, so the only way the calculate a volume of gas occupied by 1 mole of oxygen is to assume standard temperature and pressure (STP). At STP 1 mole of gas occupies 22.4 liters.

1 mole of O2 will be 22.4 liters.

0.55 moles of O2 would be:

(0.55 mole O2)*(22.4 liters) = 12.32 liters