From 23 grams of ethyl alcohol, 30 grams of sodium ethylate was obtained.

From 23 grams of ethyl alcohol, 30 grams of sodium ethylate was obtained. How many percent is this of the theoretically possible output?

To solve, we write down the equation, arrange the coefficients:
2C2H5OH + 2Na = 2C2H5ONa + H2 – substitution reaction, sodium ethylate and hydrogen were obtained;
Let’s carry out calculations according to the formulas:
M (C2H5OH) = 46 g / mol;
M (C2H5ONa) = 67.9 g / mol;
Y (C2H5OH) = m / M = 23/46 = 0.5 mol.
According to the reaction equation, the number of moles of alcohol and sodium ethylate are equal to – 2 mol, which means that Y (C2H5ONa) = 0.5 mol;
Let’s calculate the theoretical mass of sodium ethylate:
m (C2H5ONa) = Y * M = 0.5 * 67.9 = 33.95 g.
We find the yield of the reaction product by the formula:
W = m (practical) / m (theoretical) * 100;
W = 30 / 33.95 * 100 = 88.37%.
Answer: the mass fraction of the output of sodium ethylate is 88.37%.