A chemist reacted 0. 2 moles sodium benzoate with 0. 25 moles of hydrochloric acid. if she generated 22 g benzoic acid, the percent yield is 90%.
Reactant + HCl ------ Product
0.2 mol 0.25 22g
mol
Number of moles of reactant = 0.2mol
Molar mass of product = 122.12g
Number of moles of product = given mass/ molar mass
= 22g/ 122.12 g
= 0.180mol
Percent yield = (number of moles of product/ number of moles of reactant) × 100
= (0.180/0.2) × 100
= 90%
Thus we concluded that the percent yield of the given solution is 90%.
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