Answer:
2071g or 2.071kg of rust (Fe3O4)
Step-by-step explanation:
Step 1:
The balanced equation for the reaction.
3Fe + 4H2O —> Fe3O4 + 4H2
Step 2:
Determination of the mass of Fe that reacted and the mass of the Fe3O4 produced from the balanced equation.
Molar Mass of Fe = 56g/mol
Mass of Fe from the balanced equation = 3 x 56 = 168g
Molar mass of Fe3O4 = (56x3) + (16x4) = 232g/mol
Mass of Fe3O4 from the balanced equation = 1 x 232 = 232g
Summary:
From the balanced equation above,
168g of Fe reacted and 232g of Fe3O4 was produced.
Step 3:
Determination of the mass of rust (Fe3O4) produced when 1.5kg ( i.e 1500g) of Fe reacted.
This is illustrated below:
From the balanced equation above,
168g of Fe reacted to produce 232g of Fe3O4.
Therefore, 1500g of Fe will react to produce = (1500x232)/168 = 2071g of Fe3O4.
From the calculations made above, 2071g or 2.071kg of rust (Fe3O4) is produced.