What mass of iron is needed to react with sulfur in order to produce 96 grams of

iron (iii) sulfide according to the following equation?

2 fe + 3 s à 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

given the reactants, h2so4 and koh, we are asked to determine the product(s) of the reaction. since h2so4 is an acid and koh is a base. this is a neutralization reaction, thus, the products are salt and water. 

so, h+ from h2so4 will react with oh- from koh forming h2o. k+ will also react with so4 2- forming k2so4. therefore, the balanced reaction will be:

h2so4 + 2koh > k2so4 + 2h2o

explanation: