Question

How do I do this? Thank you so much for the help. I have numerous questions on my homework like this one so steps are much appreciated!

When solutions of silver nitrate, AgNO₃, and calcium iodide, CaI₂, are mixed, a yellow precipitate of silver iodide is formed. Calculate the mass of silver iodide that is formed if 50.00 mL of 1.00 M AgNO₃ is combined with 30.00 mL of 1.25 M CaI₂.

Answer 1

volume of AgNO3, V = 50.00 mL

= 5*10^-2 L

we have below equation to be used:

number of mol in AgNO3,

n = Molarity * Volume

= 1*0.05

= 5*10^-2 mol

volume of CaI2, V = 30.00 mL

= 3*10^-2 L

we have below equation to be used:

number of mol in CaI2,

n = Molarity * Volume

= 1.25*0.03

= 3.75*10^-2 mol

we have the Balanced chemical equation as:

2 AgNO3 + CaI2 ---> 2 AgI + Ca(NO3)2

2 mol of AgNO3 reacts with 1 mol of CaI2

for 5*10^-2 mol of AgNO3, 2.5*10^-2 mol of CaI2 is required

But we have 3.75*10^-2 mol of CaI2

so, AgNO3 is limiting reagent

we will use AgNO3 in further calculation

Molar mass of AgI = 1*MM(Ag) + 1*MM(I)

= 1*107.9 + 1*126.9

= 234.8 g/mol

From balanced chemical reaction, we see that

when 2 mol of AgNO3 reacts, 2 mol of AgI is formed

mol of AgI formed = (2/2)* moles of AgNO3

= (2/2)*5*10^-2

= 5*10^-2 mol

we have below equation to be used:

mass of AgI = number of mol * molar mass

= 5*10^-2*2.348*10^2

= 11.7 g

Answer: 11.7 g

Feel free to comment below if you have any doubts or if this answer do not work. I will edit it if you let me know