Cyclobutane with a mass of 14 g reacts with hydrogen to form butane

Cyclobutane with a mass of 14 g reacts with hydrogen to form butane. Determine the practical yield of butane if the mass fraction of the yield is 80% of the theoretically possible.

To solve the problem, we write the equation:
С4Н8 + Н2 = С4Н10 – hydrogenation reaction, butane was obtained;
1 mol 1 mol;
Let’s calculate the molar masses of substances:
M (C4H8) = 56g / mol;
M (C4H10) = 58g / mol;
Let us determine the number of moles of cyclobutane, if the mass is known:
Y (C4H8) = m / M = 14/56 = 0.25 mol;
According to the reaction equation, the number of moles of cyclobutane and butane is 1 mole, which means that Y (C4H10) = 0.25 mole.
We find the theoretical mass of butane:
M (C4H10) = Y * M = 0.25 * 58 = 14.5 g.
The practical mass of butane is determined by the formula:
W = m (practical) / m (theoretical) * 100;
m (practical) = 0.80 * 14.5 = 11.6 g.
Answer: the mass of butane is 11.6 g.