# 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%.