#a. 162.578 g of Fe₂S₃
#b. 7.20 g of Aluminium
#a. we are given;
131 grams of iron
Required to determine the amount of iron (iii) sulfide produced.
The equation for the reaction is;
2Fe(s) + 3S(s) → Fe₂S₃(s)
We are given 131 g of Fe and excess sulfur, this means Fe is the rate limiting reagent.
Moles of Fe that reacted:
Molar mass of Fe is 55.845 g/mol
But, moles = Mass ÷ Molar mass
Moles of Fe = 131 g ÷ 55.854 g/mol
= 2.346 moles.
But, from the equation 2 moles of Fe reacts with sulfur to produce 1 mole of iron sulfide.
Therefore;
Moles of Fe₂S₃ = moles of Fe ÷ 3
= 2.346 moles ÷ 3
= 0.782 moles
But, molar mass of Fe₂S₃ = 207.9 g/mol
Therefore;
Mass of Fe₂S₃ = 0.782 moles × 207.9 g/mol
= 162.578 g
#b. We are given;
Mass of Cu as 25.5 g
We are required to determine the mass of Al that would displace when mixed with CuSO₄
The equation for the reaction;
2Al(s) + 3CuSO₄(aq) → Al₂(SO₄)₃ + 3Cu(s)
We first determine, the moles of Cu in 25 g
Moles = Mass ÷ R.A.M
R.A.M of Cu = 63.546
Thus;
Moles of Cu = 25.5 g ÷ 63.546
= 0.401 moles
From the equation, 2 moles of Al reacts to produce 3 moles of Cu
Therefore;
Moles of Al = Moles Cu ÷ 2/3
= 0.401 moles ÷ 2/3
= 0.267 moles
But, R.A.M. of Al = 26.982
Therefore;
Mass of Al = 0.267 moles × 26.982
= 7.20 g
Thus, mass of Al required to replace 25.5 g of Cu is 7.20 g
