How much hydrogen is required to completely hydrogenate 90 g of propyne?

Given:
m (C3H4) = 90 g

To find:
n (H2) -?
m (H2) -?
V (H2) -?

Solution:
1) C3H4 + 2H2 => C3H8;
2) M (C3H4) = Mr (C3H4) = Ar (C) * N (C) + Ar (H) * N (H) = 12 * 3 + 1 * 4 = 40 g / mol;
M (H2) = Mr (H2) = Ar (H) * N (H) = 1 * 2 = 2 g / mol;
3) n (C3H4) = m (C3H4) / M (C3H4) = 90/40 = 2.25 mol;
4) n (H2) = n (C3H4) * 2 = 2.25 * 2 = 4.5 mol;
5) m (H2) = n (H2) * M (H2) = 4.5 * 2 = 9 g;
6) V (H2) = n (H2) * Vm = 4.5 * 22.4 = 100.8 liters.

Answer: The amount of substance H2 is 4.5 mol; weight – 9 g; volume – 100.8 liters.