# Question #af4de

Mar 29, 2017

$26.25 m l$

#### Explanation:

$C a C {l}_{2} + 2 A g N {O}_{3} \to C a {\left(N {O}_{3}\right)}_{2} + 2 A g C l$

From the balanced chemical equation, we can see that for every 1 unit of $C a C {l}_{2}$, 2 units of $A g N {O}_{3}$ is needed to completely react.

$\text{No. of mol of "AgNO_3} = 0.600 \cdot 0.035 = 0.021 m o l$

Hence, no. of mol of $C a C {l}_{2}$ needed to react fully is $\frac{0.021}{2} = 0.0105 m o l$

$\text{Vol. of "CaCl_2} = \frac{0.0105}{0.400} = 0.02625 l = 26.25 m l$