How much hydrogen will be released when 10 g of magnesium is treated with a solution containing 28 g of sulfuric acid?

Given:

m (Mg) = 10 g

m (H2SO4) = 28 g

To find:

V (H2) -?

Solution:

1) We compose the reaction equation according to the condition of the problem:

Mg + H2SO4 = MgSO4 + H2;

2) Find the amount of magnesium and sulfuric acid:

n (Mg) = m: M = 10 g: 24 g / mol = 0.42 mol

n (H2SO4) = 28 g: 98 g / mol = 0.286 mol

We start from a lower value to get more accurate calculations. We work with H2SO4:

3) We compose a logical expression:

if 1 mol H2SO4 gives 1 mol H2,

then 0.286 mol H2SO4 will give x mol H2,

then x = 0.286 mol.

4) Find the volume of hydrogen released during the reaction:

V (H2) = n * Vm = 0.286 mol * 22.4 L / mol = 6.4064 L;

Answer: V (H2) = 6.4064 l.