What is the volume of hydrogen gas formed (measured at room temperature and pressure)
when 0.23 g of pure sodium reacts completely with an excess of water?
(Ar values: H = 1.0; Na = 23.
Assume that the molar volume of gas at room temperature and pressure is 24 dm3.)

Answer:

26) B

Let us first write down the reaction equation.

2Na + 2H2O  –>  2NaOH + H₂

We have the Mr of Sodium = 23 and its mass = 0.23g

Moles = Mass / Mr = (0.23 / 23) = 0.01 mol

Molar ratio (Na : H) = 2 : 1

So moles of hydrogen gas produced = 0.01 / 2 = 0.005 mol

Volume of 1 mole of gas at room temperature = 24 dm³

So volume of 0.005 mol of gas = (24 x 0.005) = 0.12 dm³