Question

How many moles of precipitate will be formed when 50.0 mL of 0.300 M AgNO₃ is reacted with excess CaI₂ in the following chemical reaction? 2 AgNO₃ (aq) + CaI₂ (aq) → 2 AgI (s) + Ca(NO₃)₂ (aq)

Answer 1

The given chemical reaction is

$$ 2 A g N O_{3(a q)}+C a I_{2(a q)} \rightarrow 2 A g I_{(s)}+C a\left(N O_{3}\right)_{2(a q)} $$

The precipitate that forms is Agl.

Since Cal2 is taken in excess, the amount of AgNO3 will determine the amount of precipitate formed.

Concentration of AgNO3 used, \(C=0.300 M=0.300\) mol/L

Volume of AgNO3 used \(=\mathrm{V}=50.0 \mathrm{~mL}=0.0500 \mathrm{~L}\)

Hence, the number of moles of AgNO3 that will react is

$$ n=C \times V=0.300 \frac{\mathrm{mol}}{\mathrm{L}} \times 0.0500 \mathrm{~L}=0.0150 \mathrm{~mol} $$

Note that according to the balanced reaction 2 moles of AgNO3 results in 2 moles of Agl.

Hence, the number of moles of Agl that will be formed from 0.0150 mol of AgNO3 is

$$ \frac{2 \text { mol } A g I}{2 \operatorname{mol} A q N O_{3}} \times 0.0150 \text { mol } A g N O_{3}=0.0150 \text { mol } A g I $$