Here:
M(AgNO3)=0.20 M
M(CaCl2)=1.5 M
V(AgNO3)=1.0 L
According to balanced reaction:
1*number of mol of AgNO3 =2*number of mol of CaCl2
1*M(AgNO3)*V(AgNO3) =2*M(CaCl2)*V(CaCl2)
1*0.2 M *1.0 L = 2*1.5M *V(CaCl2)
V(CaCl2) = 0.0667 L
Answer: 0.067 L
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