Question

When 250. mL of a 0.15 M solution of ammonium sulfide (NH₄)₂S is poured into 120. mL of a 0.053 M solution of cadmium sulfate CdSO₄, how many grams of a yellow precipitate of cadmium sulfide CdS are formed? The other product is (NH₄)₂SO₄. (Hint: Write out and balance the equation. Is this a limiting reagent problem? )

(a) 5.4 g

(b) 0.92 g

(c) 2.6 g

(d) 1.9 g

(e) 530 g

Please show all steps to solution as clearly as possible please. Thank you

Answer 1

volume of (NH4)2S, V = 250 mL

= 0.25 L

we have below equation to be used:

number of mol in (NH4)2S,

n = Molarity * Volume

= 0.15*0.25

= 3.75*10^-2 mol

volume of CdSO4, V = 120 mL

= 0.12 L

we have below equation to be used:

number of mol in CdSO4,

n = Molarity * Volume

= 0.053*0.12

= 6.36*10^-3 mol

we have the Balanced chemical equation as:

(NH4)2S + CdSO4 ---> CdS + (NH4)2SO4

1 mol of (NH4)2S reacts with 1 mol of CdSO4

for 3.75*10^-2 mol of (NH4)2S, 3.75*10^-2 mol of CdSO4 is required

But we have 6.36*10^-3 mol of CdSO4

so, CdSO4 is limiting reagent

we will use CdSO4 in further calculation

Molar mass of CdS = 1*MM(Cd) + 1*MM(S)

= 1*112.4 + 1*32.07

= 144.47 g/mol

From balanced chemical reaction, we see that

when 1 mol of CdSO4 reacts, 1 mol of CdS is formed

mol of CdS formed = (1/1)* moles of CdSO4

= (1/1)*6.36*10^-3

= 6.36*10^-3 mol

we have below equation to be used:

mass of CdS = number of mol * molar mass

= 6.36*10^-3*1.445*10^2

= 0.92 g

Answer: b