Calculate the mass of the substance obtained by alcoholic fermentation of 92 g of glucose.

1) Let’s write the equation of the reaction of what happens, arrange the coefficients:
C6H12O6 = 2C2H5OH + 2CO2;
Thus, the substance ethanol (alcohol) was obtained.
M (C6H12O6) = 6Ar (C) + 12Ar (H) + 6Ar (O);
M (C6H12O6) = 6 * 12 + 1 * 12 + 6 * 16 = 72 + 12 +96 = 180 (g / mol);
M (C2H5OH) = 2Ar (C) + 5Ar (H) + Ar (O) + Ar (H);
M (C2H5OH) = 12 * 2 + 5 * 1 + 16 + 1 = 46 (g / mol);
m = n * M;
m (C2H5OH) = 2 * 46 = 92.
2) This reaction corresponds to the ratio:
92: 180 = x: 92, x is the mass of alcohol that turned out.
x = (92 * 92) / 180 = 47 (g)
Answer: the mass of alcohol obtained by alcoholic fermentation of 92 g of glucose is 47 g.

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