Question

How many moles of AgI will be formed when 70.0 mL of 0.500 M AgNO₃ is completely reacted according to the balanced chemical reaction:

2AgNO₃ + CaI₂ --> 2AgI + Ca(NO₃)₂

Answer 1

The given balanced reaction is

$2AgNO_3 + CaI_2 \rightarrow 2AgI + Ca(NO_3)_2$

Hence, two moles of AgNO3 reacts completely to give 2 moles of AgI.

We have 0.500 M AgNO3 solution with total volume of 70.0 mL

V = 70.0 mL = 0.0700 L

C = 0.500 M = 0.500 mol/L

Hence, number of moles of AgNO3 that reacts is

$C\times V = \frac{0.500 \ mol}{\cancel{L}} \times 0.0700 \ \cancel{L} = 0.0350 \ mol$

Hence, a total o 0.0350 mol of AgNO3 is reacting. Hence, the number of moles of AgI that forms is

$\frac{2 \ mol \ AgI}{2 \ \cancel{mol \ AgNO_3}} \times 0.0350 \ \cancel{mol AgNO_3} = 0.0350 \ mol \ AgI$

Hence, the number of moles of AgI formed will be 0.0350 mol.