The volume of hydrogen liberated at STP by treating 2.4 g of magnesium with excess of hydrochloric acid is _________ × 10⁻² L
Given : Molar volume of gas is 22.4 L at STP.
Molar mass of magnesium is 24 g mol⁻¹

Solution

Chemistry ma’am · 4 steps

One mole of Mg gives one mole of H₂, so find the moles of Mg and you have the gas.

1. Balanced equation. Magnesium displaces hydrogen from the acid. The mole ratio of Mg to H₂ is 1 : 1.

   Mg + 2HCl → MgCl₂ + H₂

2. Moles of Mg. n(Mg) = (2.4 g)/(24 g mol⁻¹) = 0.1 mol. The acid is in excess, so all of it reacts and gives 0.1 mol of H₂.
3. Volume at STP. V = 0.1 × 22.4 L = 2.24 L.
4. So. 2.24 L = 224 × 10⁻² L, so the blank is 224.

The common slip: 2.24. Entering the volume in litres gives 2.24, but the blank is in units of 10⁻² L, so it must be 224.

The idea to keep. Grams to moles, moles to moles with the balanced equation, then moles to litres. And always read the units in the blank before you write the number.

Answer from the NTA final answer key. Solution by PrepWiser.