Question

2 attempts left Check my work Enter your answer in the provided box. If 30.0 mL of 0.150 M CaCl, is added to 36.0 mL of 0.100 M AgNO,, what is the mass of the AgCI precipitate? 0.22

Answer 1

volume of CaCl2, V = 30 mL

= 3*10^-2 L

use:

number of mol in CaCl2,

n = Molarity * Volume

= 0.15*3*10^-2

= 4.5*10^-3 mol

volume of AgNO3, V = 36 mL

= 3.6*10^-2 L

use:

number of mol in AgNO3,

n = Molarity * Volume

= 0.1*3.6*10^-2

= 3.6*10^-3 mol

Balanced chemical equation is:

CaCl2 + 2 AgNO3 ---> 2 AgCl + Ca(NO3)2

1 mol of CaCl2 reacts with 2 mol of AgNO3

for 4.5*10^-3 mol of CaCl2, 9*10^-3 mol of AgNO3 is required

But we have 3.6*10^-3 mol of AgNO3

so, AgNO3 is limiting reagent

we will use AgNO3 in further calculation

Molar mass of AgCl,

MM = 1*MM(Ag) + 1*MM(Cl)

= 1*107.9 + 1*35.45

= 143.35 g/mol

According to balanced equation

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

= (2/2)*3.6*10^-3

= 3.6*10^-3 mol

use:

mass of AgCl = number of mol * molar mass

= 3.6*10^-3*1.434*10^2

= 0.5161 g

Answer: 0.516 g