n(CaBr2)= V(CaBr2)×c(CaBr2)=0.015×0.35=0.00525 mol
n(AgBr)=2×n(CaBr2)=2×0.00525=0.0105 mol
m(AgBr)=n(AgBr)×M(AgBr)=0.0105×188=1.974 g.
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