What volume of methanol (density 0.8) is required to obtain 37 g of methyl ester of acetic acid, if its yield is 80% of theoretical.
M (CH3COOCH3) = 74 g / mol.
37 g – 80%
X – 100%,
X = (37g × 100%): 80% = 46.25g
Let’s find the amount of CH3COOSH3 substance by the formula:
n = m: M.
n = 46.25 g: 74 g / mol = 0.625 mol.
Let’s compose the reaction equation, find the quantitative ratios of substances.
CH3OH + CH3COOH → CH3COOCH3 + H2O.
According to the reaction equation, there is CH3OH for 1 mol of CH3COOCH3. Substances are in quantitative ratios 1: 1.
The amount of substance will be the same.
n (CH3OH) = n (CH3COOCH3) = 0.625 mol.
Let’s find the mass of CH3OH by the formula:
m = n × M,
M (CH3OH) = 32 g / mol.
m = 0.625 mol × 32 g / mol = 20 g.
V = m: p.
V = 20 g: 0.8 = 25 ml.
Answer: 25 ml.
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