From 138 g of ethanol, 1.3 butadiene with a mass of 72.9 g was obtained, find the mass fraction

From 138 g of ethanol, 1.3 butadiene with a mass of 72.9 g was obtained, find the mass fraction of the yield of the reaction product.

Let’s carry out the solution, compose the reaction equation:
2С2Н5ОН = Н2С = СН – СН = СН2 + 2Н2О + Н2 – 1,3-butadiene, hydrogen and water were obtained during the reaction;
M (C2H5OH) = 46 g / mol;
M (C4H6) = 54 g / mol;
Let’s calculate the number of moles of ethanol, if the mass is known:
Y (C2H5OH) = m / M = 138/46 = 3 mol;
Let’s make the proportion:
3 mol (C2H5OH) – X mol (C4H6);
-2 mol -1 mol hence, X mol (C4H6) = 3 * 1/2 = 1.5 mol;
Determine the theoretical mass of butadiene -1.3:
m (C4H6) = Y * M = 1.5 * 54 = 81 g.
We find the mass fraction of the output of butadiene:
W = m (practical) / m (theoretical) * 100;
W = 72.9 / 81 * 100 = 90%.
Answer: the mass fraction of the butadiene yield is 90%.