Calculate the total mass of aluminum, with 90% purity, required to produce 448g of iron in the process
Fe2O3 + 2 Al = Al2O3 + 2 Fe
molesAl*1/.9=molesFe
or
massAl/molemassAl*1/.9=massFe/molemassFe
MassAl=1.11MassFe*molemassAl/molemassFe
MassAl=1.11*448*26.9/55.8 check that.
Iron must be wicked expensive where you live.
To calculate the total mass of aluminum required to produce 448g of iron, we need to use the balanced chemical equation and the concept of stoichiometry.
The balanced chemical equation for the reaction is:
Fe2O3 + 2 Al = Al2O3 + 2 Fe
From the equation, we can see that 2 moles of aluminum react with 1 mole of iron to produce 2 moles of iron oxide (Al2O3) and 2 moles of iron (Fe).
The molar mass of iron (Fe) is 55.85 g/mol.
So, the molar mass of 448g of iron is:
448 g × (1 mol/55.85 g) = 8.02 mol
According to the stoichiometry of the balanced equation, for every 2 moles of iron, we need 2 moles of aluminum. Therefore, for 8.02 moles of iron, we need:
8.02 mol × (2 mol Al/2 mol Fe) = 8.02 mol of aluminum.
Since we are provided with the purity of aluminum (90%), we can calculate the mass of aluminum required by multiplying the molar mass of aluminum (26.98 g/mol) with the number of moles:
8.02 mol × (26.98 g/mol) = 216.10 g
So, the total mass of aluminum required is 216.10 grams.