What volume of gas will be released when 6.5 g of zinc react with 19.6 g of sulfuric acid?

Given:
m (Zn) = 6.5 g
m (H2SO4) = 19.6 g
To find:
V (H2) =?
Decision:
1) Let’s compose the equation of the chemical reaction, let the volume of hydrogen be x l.
Zn + H2SO4 = ZnSO4 + H2
2) Let’s find the substance by which we will count, for this we will find the amount of each.
n = m / M
M (Zn) = 65 g / mol
n (Zn) = 6.5 g: 65 g / mol = 0.1 mol
M (H2SO4) = 2 + 32 + 64 = 98 g / mol
n (H2SO4) = 19.6 g: 98 g / mol = 0.2 mol
The calculation is carried out by lack, which means we will count by zinc.
3) Find the volume of hydrogen.
6.5 g: 65 g / mol = x L: 22.4 L / mol
x = (6.5 x 22.4): 65 = 2.24 liters.
Answer: V (H2) = 2.24 liters.