Determine the mass of glycerin obtained from 190 grams of fat containing 80% glyceride

| September 10, 2021 | education

Determine the mass of glycerin obtained from 190 grams of fat containing 80% glyceride tristearate, assuming a 93% yield.

Given:
m (fat) = 190 g
ω ((C17H35COO) 3C3H5) = 80%
ω out. = 93%

Find:
m pract. (C3H5 (OH) 3) -?

Solution:
1) (C17H35COO) 3C3H5 + 3NaOH => 3C17H35COONa + C3H5 (OH) 3;
2) M ((C17H35COO) 3C3H5) = Mr ((C17H35COO) 3C3H5) = Ar (C) * 54 + Ar (H) * 110 + Ar (O) * 6 = 12 * 57 + 1 * 110 + 16 * 6 = 890 g / mol;
M (C3H5 (OH) 3) = Mr (C3H5 (OH) 3) = Ar (C) * 3 + Ar (H) * 8 + Ar (O) * 3 = 12 * 3 + 1 * 8 + 16 * 3 = 92 g / mol;
3) m ((C17H35COO) 3C3H5) = ω ((C17H35COO) 3C3H5) * m (fat) / 100% = 80% * 190/100% = 152 g;
4) n ((C17H35COO) 3C3H5) = m ((C17H35COO) 3C3H5) / M ((C17H35COO) 3C3H5) = 152/890 = 0.17 mol;
5) n (C3H5 (OH) 3) = n ((C17H35COO) 3C3H5) = 0.17 mol;
6) m theor. (C3H5 (OH) 3) = n (C3H5 (OH) 3) * M (C3H5 (OH) 3) = 0.17 * 92 = 15.64 g;
7) m practical. (C3H5 (OH) 3) = ω out. * m theor. (C3H5 (OH) 3) / 100% = 93% * 15.64 / 100% = 14.55 g.

Answer: The practical mass of C3H5 (OH) 3 is 14.55 g.



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