Find the mass of glucose that must be fermented to obtain ethanol 5.2 g, if the yield

Find the mass of glucose that must be fermented to obtain ethanol 5.2 g, if the yield of the reaction product from theoretically possible is 80%

Let’s find the mass of С2Н5ОН.

5.2 g – 80%,

X – 100%,

X = (5.2 × 100%): 80% = 6.5 g.

Let’s find the amount of ethanol substance.

M (C2H5OH) = 46 g / mol.

n = 6.5 g: 46 g / mol = 0.14.

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

С6Н12О6 → 2 С2Н5ОН + 2СО2 ↑.

According to the reaction equation, for 2 mol of С2Н5ОН there is 1 mol of С6Н12О6. Substances are in quantitative ratios of 2: 1.

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

n (C6H12O6) = ½ n (C2H5OH) = 0.14: 2 = 0.7 mol.

Let’s find the mass of С6Н12О6.

m = n × M,

M (C6H12O6) = 180 g / mol.

m = 0.7 mol × 180 g / mol = 126 g.