what mass of hematite (fe^2O^3)is required to react with excess carbon and produce 2.845g of iron, if the % yield is 75.369
formula is 2fe^2O^3+3C=4Fe+3CO2
How many moles is 2.845 g of Fe?
then moles of Fe2O3 must be molesFe/.75369 .
Check my thinking.
To find the mass of hematite required, we need to use the balanced equation and the given information.
Step 1: Write and balance the equation:
2Fe2O3 + 3C -> 4Fe + 3CO2
The equation shows that two moles of Fe2O3 react with three moles of C to produce four moles of Fe.
Step 2: Calculate the molar mass of Fe2O3:
Fe2O3 has a molar mass of (2 x molar mass of Fe) + (3 x molar mass of O)
The molar mass of Fe is 55.845 g/mol, and the molar mass of O is 16.00 g/mol.
Thus, the molar mass of Fe2O3 is (2 x 55.845) + (3 x 16.00) = 159.69 g/mol.
Step 3: Calculate the moles of Fe produced:
Given mass of Fe = 2.845 g
Molar mass of Fe = 55.845 g/mol
Moles of Fe = (given mass of Fe) / (molar mass of Fe)
Moles of Fe = 2.845 g / 55.845 g/mol = 0.0509 mol
Step 4: Calculate the theoretical yield:
According to the balanced equation, 2 moles of Fe2O3 produce 4 moles of Fe.
Therefore, the stoichiometric ratio is 2:4, which simplifies to 1:2.
Moles of Fe2O3 required = (moles of Fe) / 2 = 0.0509 mol / 2 = 0.0254 mol
Step 5: Calculate the mass of Fe2O3 required:
Mass of Fe2O3 required = (moles of Fe2O3 required) x (molar mass of Fe2O3)
Mass of Fe2O3 required = 0.0254 mol x 159.69 g/mol = 4.068 g
However, we are given that the percent yield is 75.369%. The percent yield is the ratio of the actual yield to the theoretical yield, expressed as a percentage.
Step 6: Calculate the actual yield:
Actual yield = (percent yield / 100) x (theoretical yield)
Actual yield = (75.369 / 100) x 4.068 g
Actual yield = 3.063 g
Therefore, the mass of hematite (Fe2O3) required to produce 2.845 g of iron with a 75.369% yield is approximately 4.068 g.