From 42 g of ethylene 3.45 g of ethanol was obtained. How many percent was this of the theoretically possible output?

To solve the problem, make up the equation:
m = 42 g. m = 3.45 g; W -?
1. Н2С = СН2 + Н2О = С2Н5ОН – hydration, ethanol was obtained;
2. Calculations:
M (C2H4) = 28 g / mol;
M (C2H5OH) = 46 g / mol.
3. Determine the amount of the original substance, if the mass is known:
Y (C2H4) = m / M = 42/28 = 1.5 mol;
Y (C2H5OH) = 1.5 mol since the amount of substances is equal to 1 mol according to the equation.
4. Find the mass and yield of the product:
m (C2H5OH) = Y * M = 1.5 * 46 = 69 g;
W = m (practical) / m (theoretical) * 100;
W = 3.45 / 69 * 100 = 5%
Answer: the ethanol yield is 5%.