Answer to Question #112979 in General Chemistry for Sabrina

Question

A student conducts an experiment by reacting 23 g K2CrO4 with an excess of Pb(NO3)2. The reaction yields 21.027 g KNO3. What is the percent yield?

K2CrO4(aq) + Pb(NO3)2(aq) → PbCrO4(s)+ 2 KNO3(aq)

Answer 1

m_teor(KNO3)=m(K2CrO4)*2*Mr(KNO3)/Mr(K2CrO4)=23*2*101/194=23.95 g

%yield=m_pract/m_teor=21.027/23.95=87.8%