Question

When the reaction occurred, the amount of Ag-S obtained was 45.0g. What is the percent yield of the reaction

Answer 1

A silver metal reacts with sulfur to form silver sulfide according to the following reaction:

2Ag(s) + S(s) → Ag₂S(s)

Identify the limiting reagent if 50.0 g Ag reacts with 10.0 g S

When the reaction occurred, the amount of Ag-S obtained was 45.0g. What is the percent yield of the reaction?

M(Ag) = 107.86 g/mol

M(S) = 32.06 g/mol

$n(Ag) = \frac{50.0}{107.86} = 0.463 \;mol \\ n(S) = \frac{10.0}{32.06} = 0.312 \;mol$

For each mole of S we need two moles of Ag, so Ag is limiting reactant.

According to the reaction:

n(Ag₂S) $= \frac{1}{2}n(Ag) = \frac{1}{2} \times 0.463 = 0.2315 \;mol$

M(Ag₂S) = 247.78 g/mol

m(Ag₂S) $= 0.2315 \times 247.78 = 57.36 \;g$

Proportion:

57.36 g – 100 %

45.0 g – x

$x = \frac{45.0 \times 100}{57.36} = 47.45 \; \%$

Answer: 47.45 %