Answer to Question #96239 in General Chemistry for Mitch

Question

1. How many grams of Cu(OH)2 will precipitate when excess Ba(OH)2 solution is added to 53.0 mL of 0.605 M CuSO4 solution? ___g

CuSO4(aq) + Ba(OH)2(aq) Cu(OH)2(s) + BaSO4(aq)

2. How many mL of 0.712 M HBr are needed to dissolve 8.59 g of CaCO3?___g

2HBr(aq) + CaCO3(s) CaBr2(aq) + H2O(l) + CO2(g)

Answer 1

1.n(CuSO4)=0.053L*0,605 М=0.0321moles

n(CuSO4)=n(Cu(OH)2)=0.0321moles

m(Cu(OH)2)=97,56(g/mol)*0.0321moles=3.132g

2.n(CaCO3)=8.59g/100(g/mol)=0.0859moles

n(HBr)=0.0859moles*2=0.1718moles

V(HBr)=0.1718moles*0.712M=0.1223L=122.3mL