What mass of iron is needed to react with sulfer in order to produce 96 grand of Fe2S3 according to this equation: 2Fe + 3S = Fe2S3?

The balanced equation for the reaction is as follows;

2Fe + 3S - --> Fe₂S₃

molar ratio of Fe to Fe₂S₃ is 2:1

mass of Fe₂S₃ to be produced is - 96 g

therefore number of moles of Fe₂S₃ to be produced is - 96 g / 208 g/mol

number of Fe₂S₃ moles = 0.46 mol

according to the molar ratio

when 2 mol of Fe reacts with 3 mol of sulfur then 1 mol of Fe₂S₃ is produced

that for 1 mol of Fe₂S₃ to be produced - 2 mol of Fe should react

therefore for 0.46 mol of Fe₂S₃ to be produced - 2 x 0.46 = 0.92 mol of Fe is required

mass of Fe required - 0.92 mol x 56 g/mol = 51.5 g

mass of Fe required is - 51.5 g

Answer: 51.69g

Explanation:

2Fe + 3S - > Fe2S3

Molar Mass of Fe2S3 = (2x56) + (32x3) = 112 + 96 = 208g/mol

Molar Mass of Fe = 56g/mol

Mass conc. Of Fe = 2x56 = 112g

From the equation,

112g of Fe produced 208g of Fe2O3.

Therefore, Xg of Fe will produce 96g of Fe2O3 i. e

Xg of Fe = (112x96) / 208 = 51.69g