90 g of glucose underwent alcoholic fermentation. The gas obtained in this case was completely absorbed by 200 g of a 20%

90 g of glucose underwent alcoholic fermentation. The gas obtained in this case was completely absorbed by 200 g of a 20% sodium hydroxide solution. The molar mass of the resulting salt?

p-equation: С6Н12О6 → 2С2Н5ОН + 2СО2 ↑
From 1 mol of glucose, 2 mol of CO2 are formed, therefore:
n (CO2) = 2n (glucose) = 2m () / M (glucose) = 2 ∙ 90/180 = 1 mol
n (NaOH) = m (p-pa) ∙ ω / М (NaOH) = 200 ∙ 0.2 / 40 = 1 mol
Those. CO2 and NaOH reacted in a 1: 1 ratio, and salt of the composition NaHCO3 was formed according to the conservation of the substance:
CO2 + NaOH → NaHCO3
M (NaHCO3) = 84 g / mol

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