How many grams of iron (3)sulfide are produced when 131 grams of iron react with excess sulfur?

how many grams of al will completely replace 25.5 grams of cu when aluminum is mixed with aqueous copper (2) sulfate?

#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

1 a

2 d

3 a

4 c

explanation: