How many moles of AgI will be formed when 75.0 mL of 0.300 M AgNO3 completely reacted according to the balanced chemical reaction:

Question

: 2AgNO₃(aq) + CaI₂(aq) → 2AgI(s) + Ca(NO₃)₂(aq)

I know I have to use conversion factors but I'm not sure how to do the problem at all.

Answer 1

Look at the balanced equation. TWO moles AgNO₃ makes TWO moles of AgI. So, they are in a 1:1 mole ratio. Simply find out how many moles of AgNO₃ you have and that will be the number of moles of AgI that can be formed.

moles AgNO₃ = (0.075 L)(0.300 mol/L) = 0.0225 moles

moles AgI = 0.0225 mol AgNO₃ x 2 mol AgI/2 mol AgNO₃ = 0.0225 moles AgI

Answer 2

Usually, I solve those types of questions in this way (another way of doing the conversion factor).

1) Let’s find the number of moles of 75 mL, 0.3M AgNO₃:

n = M*VL = 0.3 * 75 * 10-³ = 0225 mol

2) From the equation:

2 mol AgNO₃ --> 2 mol of AgI

0.0225 mol AgNO₃ --> x

x = (2*0.0225)/2 = 0.0225 mol of AgI