12 grams of benzene was obtained from 13.44 liters of acetylene. What percentage is the theoretical output.

Given:

V (C2H2) = 13.44 L

m (C6H6) = 12 g

To find:

w% (C6H6 practical) -?

Decision:

3C2H2 — (act. C) -> C6H6, – we solve the problem, relying on the composed reaction equation:

1) Find the amount of acetylene required for the reaction:

n (C2H2) = V: Vm = 13.44 L: 22.4 L / mol = 0.6 mol

2) We compose a logical expression:

if 3 mol C2H2 gives 1 mol C6H6,

then 0.6 mol C2H2 will give x mol C6H6,

then x = 0.2 mol.

3) Find the mass of benzene obtained during the reaction:

m (C6H6) = n * M = 0.2 mol * 78 g / mol = 15.6 g

4) Find the percentage yield of benzene from the theoretical:

w% (C6H6 practical) = 12 g: 15.6 g * 100% = 77%.

Answer: w% (C6H6 practical) = 77%.