Question

How do I solve this?

Answer 1

Explanation:

a) Since this is a double displacement reaction, we write the balanced equation as

$2AgNO_3(aq) + CaCl_2(aq) \\ \rightarrow 2AgCl(s) + Ca(NO_3)_2(aq)$

b) Next we find the number of moles of AgNO3 in the solution.

$(0.005\:\text{L})(0.500\:M\:AgNO_3) \\ = 0.0025\:\text{mol}\:AgNO_3$

Next, use the molar ratio to find the necessary amount of CaCl2 to react with the AgNO3:

$0.0025\:\text{mol}\:AgNO_3× \left(\dfrac{1\:\text{mol}\:CaCl_2}{2\:\text{mol}\:AgNO_3} \right)$

$= 0.00125\:\text{mol}\:CaCl_2$

The volume of 0.500 M solution of CaCl2 necessary to react all of the given AgNO_3 is then

$V = \dfrac{0.00125\:\text{mol}\:CaCl_2}{0.500\:\text{M}\:CaCl_2}$

$= 0.0025\:\text{L} = 2.5\:\text{mL}\:CaCl_2$

c) The theoretical yield can then be calculated as

$0.0025\:\text{mol}\:AgNO_3 × \left(\dfrac{2\:\text{mol}\:AgCl}{2\:\text{mol}\:AgNO_3} \right)$

$= 0.0025\:\text{mol}\:AgCl$

Converting this amount of AgCl into grams, we get

$0.0025\:\text{mol}\:AgCl × \left(\dfrac{143.32\:\text{g}\:AgCl}{1\:\text{mol}\:AgCl} \right)$

$= 0.358\:\text{g}\:AgCl$