What volume of hydrogen will be released when potassium interacts with sulfate acid with an amount

What volume of hydrogen will be released when potassium interacts with sulfate acid with an amount of 5 mol of a substance?

1. Let’s write the equation of interaction of potassium with sulfuric acid:

2K + H2SO4 = K2SO4 + H2 ↑;

2.In the reaction equation, there are unit coefficients in front of the acid and the evolving hydrogen, so the amount of hydrogen is:

n (H2) = n (H2SO4) = 5 mol;

3.Calculate the volume of hydrogen:

V (H2) = n (H2) * Vm = 5 * 22.4 = 112 HP

Answer: 112 liters.