210 g of methylcyclopropane were treated with 118 liters of hydrogen. Find the mass of the resulting product.

Let’s implement the solution:

According to the condition of the problem, we write down the equation of the process:
m = 210 g X g -?

CH2 – CH2 – CH – (CH3) + H2 = CH3 – CH – (CH3) – CH3 – the cycle is destroyed when heated to 80 degrees, the catalyst is Ni, hydrogen is added, 2 – methyl propane is obtained;

Calculations:
M (methylcyclopropane) = 56 g / mol;

M (2-methyl propane) = 58 g / mol.

Determine the amount of the original substance:
Y (methylcyclopropane) = m / M = 210/56 = 3.75 mol (deficient substance).

Calculations are made taking into account the deficient substance.

Y (2-methyl propane) = 3.75 mol since the amount of these substances is 1 mol.

Proportion:
1 mol of gas at normal level – 22.4 liters;

X mol (H2) -118 L from here, X mol (H2) = 1 * 118 / 22.4 = 5.26 mol (substance in excess).

Find the mass of the product:
m (2-methyl propane) = Y * M = 3.75 * 58 = 217.5 g

Answer: the mass of 2-methyl propane is 217.5 g