Question

How many grams of AgBr will form when 45.0 mL of 0.842 M CaBr2 react?

CaBr2(aq) + 2 AgNO3(aq) → 2 AgBr(s) + Ca(NO3)2(aq)

Answer 1

$45mL = 0.045L$

$n = C*V$

$n(CaBr_2)=0.842*0.045mol$

$n(AgBr)= 2*n(CaBr_2)=$

$=2*0.842*0.045mol=0.07578mol$

$m=n*M$

$M(AgBr)= 188g/mol$

$m(AgBr)= 188*0.07578g=14.24664g$

Answer: $m(AgBr)=14.24664g$