What mass of alcohol can be obtained as a result of processing 12 kg of glucose, if the reaction yield is 80%

Let’s find the amount of the substance С6Н12О6.

n = m: M.

M (C6H12O6) = 180 kg / kmol.

n = 12 kg: 180 kg / kmol = 0.067 kmol.

Let’s compose the reaction equation. Let’s find the quantitative ratios of substances.

С6Н12О6 → 2 С2Н5ОН + 2СО2 ↑.

According to the reaction equation, there is 2 kmol of C2H5OH per 1 kmol of С6Н12О6. Substances are in quantitative ratios of 1: 2.

The amount of C2H5OH substance is 2 times more than C6H12O6.

n (C2H5OH) = 2n (C6H12O6) = 0.067 × 2 = 0.134 kmol.

Let’s find the mass of С2Н5ОН.

M (C2H5OH) = 46 kg / kmol.

m = n × M.

m = 46 kg / kmol × 0.134 kmol = 6.164 kg.

6.164 kg – 100%,

X – 80%,

X = (6.164 × 80%): 100% = 4.93 kg.

Answer: 4.93 kg.