Question

100. mL of 0.200 M AgNO3(aq) is mixed with 30 mL of 0.100 M Na2S(aq). Calculate the mass (in g) of any precipitate that is formed. Enter your answer to 2 decimal places or if no precipitate forms.

Answer 1

volume of AgNO3, V = 1*10^2 mL

= 0.1 L

use:

number of mol in AgNO3,

n = Molarity * Volume

= 0.2*0.1

= 2*10^-2 mol

volume of Na2S, V = 30 mL

= 3*10^-2 L

use:

number of mol in Na2S,

n = Molarity * Volume

= 0.1*3*10^-2

= 3*10^-3 mol

Balanced chemical equation is:

2 AgNO3 + Na2S ---> Ag2S + 2 NaNO3

2 mol of AgNO3 reacts with 1 mol of Na2S

for 2*10^-2 mol of AgNO3, 1*10^-2 mol of Na2S is required

But we have 3*10^-3 mol of Na2S

so, Na2S is limiting reagent

we will use Na2S in further calculation

Molar mass of Ag2S,

MM = 2*MM(Ag) + 1*MM(S)

= 2*107.9 + 1*32.07

= 247.87 g/mol

According to balanced equation

mol of Ag2S formed = (1/1)* moles of Na2S

= (1/1)*3*10^-3

= 3*10^-3 mol

use:

mass of Ag2S = number of mol * molar mass

= 3*10^-3*2.479*10^2

= 0.7436 g

Answer: 0.74 g