Answer :
The theoretical yield of iron (55.845 g/mol) when 10.0 g Fe₂O₃ (159.69 g/mol) reacts with excess carbon (12.01 g/mol) is 6.99 g.
the balance equation is given as :
Fe₂O₃ + 3C ----> 2Fe + 3CO
the mass of Fe₂O₃ = 10 g
molar mass of Fe₂O₃ = 159.69 g / mol
the number of moles of Fe₂O₃ = mass / molar mass
= 10 / 159.69
= 0.062 mol
1 mole of Fe₂O₃ produce 2 mole of Fe
0.062 mol of Fe₂O₃ = 2 × 0.062 = 0.125 mol of Fe
mole of Fe = 0.125 mol
molar mass of Fe = 55.8 g/mol
mass of Fe = 0.125 × 55.5
= 6.99 g
The theoretical yield = 6.99 g
Thus, The theoretical yield of iron (55.845 g/mol) when 10.0 g Fe₂O₃ (159.69 g/mol) reacts with excess carbon (12.01 g/mol) is 6.99 g.
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