For the reaction AgNO3 (aq) + KBr AgBr precipitates out.?

Question

Calculate the percent yield (%) if 15.953 g of AgBr precipitates
when 2.250 L of 0.0600 M AgNO3 (aq) and 2.240 L of 0.0464 M KBr are mixed

Answer 1

the moles of AgNO3 : 2.25 * 0.06 = 0.135
the moles of KBr : 2.24 * 0.0464 = 0.103936
the ratio of KBr < AgNO3 => KBr reacts completely
the moles of AgNO3 produced with 100% : 0.103936
the grams of AgNO3 0.103936* 170 = 17.669 g
the percent yield 15.953 / 17.669 * 100% = 90.29%