Answer:
59.9% is the percent yield for the 4.5 g of produced Al₂S₃
Step-by-step explanation:
Let's determine the reaction:
3S + 2Al → Al₂S₃
First of all, let's determine the limiting reactant. We need to convert the mass to moles:
4.8 g /32.06g/mol = 0.150 moles of S
5.4 g / 26.98 g/mol = 0.200 moles of Al
3 moles of S react to 2 moles of Al
Then, 0.150 moles of S may react to (0.150 . 2)/3 = 0.1 ,moles of Al
We have 0.200 moles and we only have 0.1. As we have excess of Al, this is the excess reactant. In conclussion, the limiting reagent is S.
2 moles of Al react to 3 moles of S
Then 0.2 moles of Al may react to (0.2 . 3) /2 = 0.3 moles of S. (We only have 0.150 moles)
Let's go to the product, 3 moles of S can produce 1 mol of Al₂S₃
Then 0.150 moles of S, may produce (0.150 . 1) /3 = 0.05 moles.
We convert moles to mass to determine the thoretical yield:
0.05 mol . 150.15g /mol = 7.50g
Percent yield = (Produced yield/Theoretical yield) . 100
% = (4.5g / 7.5g) . 100 = 59.9%