From 84 liters of butane with a volume fraction of impurities of 20% by two-stage synthesis, butanol-2 with

From 84 liters of butane with a volume fraction of impurities of 20% by two-stage synthesis, butanol-2 with a mass of 55.5 g was obtained. The proportions of the yield of the products of the first and second stages of synthesis are the same. Find the yield rates of the products.

Let’s implement the solution:
1. In accordance with the condition, we compose the equations:
1st stage. Н3С – СН2 – СН2 – СН3 = Н2 + Н2С = СН – СН2 – СН3 – dehydrogenation, butene -1 was obtained;
2nd stage. H2C = CH – CH2 – CH3 + H2O = H3C – CH (OH) – CH2 – CH3 – hydration, butanol -1 was formed;
2. Calculations:
M (C4H8) = 56 g / mol.
M (C4H9OH) = 74 g / mol.
3. Determine the volume of the original substance without impurities:
V (C4H10) = 84 * (1 – 0.20) = 67.2 liters.
4. Proportion:
1 mol at normal level – 22.4 l;
X mol (C4H10) – 67.2 liters. hence, X mol (C4H10) = 1 * 67.2 / 22.4 = 3 mol.
Y (C4H8) = 3 mol since the amount of substances according to the equation is 1 mol.
Y (C4H9OH) = 3 mol.
5. Find the mass and yield of the product:
m (C4H9OH) = Y * M = 3 * 74 = 222 g;
W = m (practical) / m (theoretical) * 100;
W = 55.5 / 222 * 100 = 25%
Answer: the mass fraction of the output of butanol – 1, butene – 1 is equal to 25%.