If the conversion of Iron ore to iron is represented by Fe2O3(s) + 3CO(g) --> 2Fe(s) + 3CO2(g) produces 147 g of iron from 500 g of iron ore, what is the percent yield?
Is the answer 29.4% iron ore?
Thanks.
Is the answer 29.4% iron ore?
Thanks.
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Is the answer 29.4% ? no,sorry, it's not
Assuming the iron ore is pure Fe2O3
Use the balanced equation.
Fe2O3(s) + 3CO(g) --> 2Fe(s) + 3CO2(g)
that tells you 1 mole Fe2O3 = 160g gives 2 moles Fe =112g
Use simple proportion
500g Fe2O3 gives 112/160 x 500= 350g
% yield = actual amount / calculated amount x 100 = 147/350 x 100 = 42%
Assuming the iron ore is pure Fe2O3
Use the balanced equation.
Fe2O3(s) + 3CO(g) --> 2Fe(s) + 3CO2(g)
that tells you 1 mole Fe2O3 = 160g gives 2 moles Fe =112g
Use simple proportion
500g Fe2O3 gives 112/160 x 500= 350g
% yield = actual amount / calculated amount x 100 = 147/350 x 100 = 42%