Question

How do I solve this?

Answer 1

Step-by-step 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(\frac{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 = \frac{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(\frac{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(\frac{143.32\:\text{g}\:AgCl}{1\:\text{mol}\:AgCl} \right)`

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