Question

Какую массу этилового спирта можно получить гидратацией 112 литров этилена, если доля выхода продукта реакция составляет 80% от теоретически возможного.

Answer 1

С2Н4+Н2О=С2Н5ОН

v(C2H4)=112/22,4=5 моль=v(C2H5OH)
m(C2H5OH)=v×M=5×46=230 г - теоретическая
практическая m(C2H5OH)=230×0,8=184 г

Answer 2

Calculation of the Mass of Ethanol Obtained by Hydration of Ethylene

To calculate the mass of ethanol obtained by the hydration of 112 liters of ethylene, we need to consider the yield of the reaction. The given information states that the yield of the reaction is 80% of the theoretically possible yield.

The molar mass of ethylene (C2H4) is 28 g/mol, and the molar mass of ethanol (C2H5OH) is 46 g/mol.

To calculate the mass of ethylene, we can use the ideal gas law equation:

PV = nRT

Where: P = pressure (assumed to be constant) V = volume of ethylene (112 liters) n = number of moles of ethylene R = ideal gas constant T = temperature (assumed to be constant)

Since the pressure and temperature are assumed to be constant, we can simplify the equation to:

V = nR

To find the number of moles of ethylene, we can rearrange the equation:

n = V / R

Substituting the values:

n = 112 L / (0.0821 L·atm/mol·K) = 1365.9 mol

Now, we can calculate the theoretical yield of ethanol by multiplying the number of moles of ethylene by the molar ratio of ethanol to ethylene. From the balanced chemical equation for the hydration of ethylene, we know that 1 mole of ethylene produces 1 mole of ethanol:

Theoretical yield of ethanol = 1365.9 mol × 1 mol ethanol / 1 mol ethylene = 1365.9 mol

To calculate the mass of ethanol, we can multiply the number of moles of ethanol by its molar mass:

Mass of ethanol = 1365.9 mol × 46 g/mol = 62,835.4 g

However, we need to consider that the yield of the reaction is only 80% of the theoretically possible yield. Therefore, the actual mass of ethanol obtained will be 80% of the calculated mass:

Actual mass of ethanol = 80% × 62,835.4 g = 50,268.32 g

Therefore, the mass of ethanol that can be obtained by the hydration of 112 liters of ethylene, with a yield of 80% of the theoretically possible yield, is approximately 50,268.32 grams.