You need to get 23 g of ethanol. The mass fraction of the alcohol yield is 90%

| August 13, 2021 | education

You need to get 23 g of ethanol. The mass fraction of the alcohol yield is 90% of the theoretically possible. What is the mass of ethylene required for the reaction?

To solve, we write down the equation of the process:

X g -? m = 23 g; W = 90%

С2Н4 + Н2О = С2Н5ОН – hydration, ethyl alcohol was obtained;

Calculations by formulas:
M (C2H4) = 28 g / mol;

M (C2H5OH) = 46 g / mol.

Taking into account the product yield, we find the mass, the amount of alcohol:
W = m (practical) / m (theoretical) * 100

m (C2H5OH) = 23 / 0.90 = 25.5 g

Y (C2H5OH) = m / M = 25.5 / 46 = 0.6 mol;

Y (C2H4) = 0.6 mol since the amount of substances is 1 mol each.

We find the mass of the original substance:
m (C2H4) = Y * M = 0.6 * 28 = 16.8 g

Answer: you need ethylene weighing 16.8 g



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