Дано
m(ppa HCL)=60 g
W(HCL)=3%
m(ppaAgNO3)=170g
W(AgNO3)=10%
----------------------
m(AgCL)-?
m(HCL)=60*3%/100%=1.8 g
m(AgNO3)=170*10%/100%=1.7 g
M(HCL)=36.5 g/mol
n(HCL)=m/M=1.8/36.5=0.049 mol
M(AgNO3)=170 g/mol
n(AgNO3)=m/M=1.7/ 170=0.01 mol
n(HCL)>n(AgNO3)
M(AgCL)=143.5 g/mol
AgNO3+HCL-->AgCL+HNO3
n(AgNO3)=n(AgCL) = 0.01 mol
m(AgCL)=n*M=0.01*143.5=1.435 g
ответ 1.435 г