How much hydrogen is required to completely hydrogenate 423 g of oleic acid?

Given:
m (C17H33COOH) = 423 g

To find:
V (H2) -?

1) C17H33COOH + H2 => C17H35COOH;
2) M (C17H33COOH) = Mr (C17H33COOH) = Ar (C) * N (C) + Ar (H) * N (H) + Ar (C) * N (C) + Ar (O) * N (O ) + Ar (O) * N (O) + Ar (H) * N (H) = 12 * 17 + 1 * 33 + 12 * 1 + 16 * 1 + 16 * 1 + 1 * 1 = 282 g / mol ;
3) n (C17H33COOH) = m (C17H33COOH) / M (C17H33COOH) = 423/282 = 1.5 mol;
4) n (H2) = n (C17H33COOH) = 1.5 mol;
5) V (H2) = n (H2) * Vm = 1.5 * 22.4 = 33.6 liters.

Answer: The volume of hydrogen is 33.6 liters.