What is the mass of lithium when interacting with sulfuric acid if 2.24 liters of hydrogen were released?

1. The interaction of lithium with sulfuric acid proceeds according to the equation:

2Li + H2SO4 = Li2SO4 + H2 ↑;

2.Calculate the chemical amount of released hydrogen:

n (H2) = V (H2): Vm = 2.24: 22.4 = 0.1 mol;

3.the stoichiometric coefficients of lithium and hydrogen in the equation are related as 2: 1, we determine the amount of lithium:

n (Li) = n (H2) * 2 = 0.1 * 2 = 0.2 mol;

4.Calculate the mass of the interacting lithium:

m (Li) = n (Li) * M (Li) = 0.2 * 7 = 1.4 g.

Answer: 1.4 g.