What volume of an alcohol solution with a mass fraction of alcohol of 96% (p = 0.8 g ml) is required to obtain 1 kg of butadiene

What volume of an alcohol solution with a mass fraction of alcohol of 96% (p = 0.8 g ml) is required to obtain 1 kg of butadiene by the Lebedev method with a reaction yield of 90% of the theoretical?

Given:
ω (C2H5OH) = 96% ρ solution (C2H5OH) = 0.8 g / ml m (C4H6) = 1 kg = 1000 g ω out. = 90%
Find: V (C2H5OH) -?
Decision:
1) Write the reaction equation: 2C2H5OH = (tC? Cat.) => C4H6 + 2H2O + H2;
2) Find the theoretical mass C4H6: m theor. (C4H6) = m practical. (C4H6) * 100% / ω out. = 1000 * 100% / 90% = 1111.11 g;
3) Find the amount of substance C4H6: n (C4H6) = m (C4H6) / Mr (C4H6) = 1111.11 / 54 = 20.58 mol;
4) Find the amount of substance C2H5OH: n (C2H5OH) = n (C4H6) * 2 = 20.58 * 2 = 41.16 mol;
5) Find the mass of C2H5OH: m (C2H5OH) = n (C2H5OH) * Mr (C2H5OH) = 41.16 * 46 = 1893.36 g;
6) Find the mass of the C2H5OH solution: m solution (C2H5OH) = m (C2H5OH) * 100% / ω (C2H5OH) = 1893.36 * 100% / 96% = 1972.25 g;
7) Find the volume of the C2H5OH solution: V solution (C2H5OH) = m solution (C2H5OH) / ρ solution (C2H5OH) = 1972.25 / 0.8 = 2465.3 ml.
Answer: The volume of the C2H5OH solution is 2465.3 ml.