Indicate the mass of alanine taken for the reaction with hydrochloric acid, if 45.18 g

Indicate the mass of alanine taken for the reaction with hydrochloric acid, if 45.18 g of salt is formed, and the yield of the reaction product is 90%.

Given:
m pract. (salt) = 45.18 g
η (salt) = 90%

Find:
m (C3H7NO2) -?

Solution:
1) C3H7NO2 + HCl => [C3H8NO2] Cl;
2) M ([C3H8NO2] Cl) = 125.5 g / mol;
M (C3H7NO2) = 89 g / mol;
3) m theor. ([C3H8NO2] Cl) = m practical. ([C3H8NO2] Cl) * 100% / η ([C3H8NO2] Cl) = 45.18 * 100% / 90% = 50.2 mol;
4) n theory. ([C3H8NO2] Cl) = m ([C3H8NO2] Cl) / M ([C3H8NO2] Cl) = 50.2 / 125.5 = 0.4 mol;
5) n (C3H7NO2) = n theory. ([C3H8NO2] Cl) = 0.4 mol;
6) m (C3H7NO2) = n (C3H7NO2) * M (C3H7NO2) = 0.4 * 89 = 35.6 g.

Answer: The mass of C3H7NO2 is 35.6 g.