Determine the mass of nitroethane formed during the nitration of 11.2 L of ethane if the product yield is 80% (26 g).

We carry out the solution:
1. In accordance with the condition, we write the equation:
V = 11.2 liters. X g -? W = 80%;
C2H6 + HNO3 = C2H5NO2 + H2O – nitration, obtained nitroethane, water;
2. Let’s calculate:
M (C2H5NO2) = 75 g / mol.
3. Proportion:
1 mol of gas at normal level – 22.4 liters;
X mol (C2H6) – 11.2 liters. hence, X mol (C2H6) = 1 * 11.2 / 22.4 = 0.5 mol.
4. Find the theoretical mass of the product:
m (C2H5NO2) = Y * M = 0.5 * 75 = 37.5 g;
m (C2H5NO2) = 0.80 * 37.5 = 30 g.
Answer: nitroethane with a mass of 30 g was obtained in the process of nitration.

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