Actual yield in moles :
1 mol Fe2O3 gives 2 mol Fe
0.322 mol Fe2O3 gives= 2x0.322/1= 0.644 mol Fe
Theoretical yield:
0.322 mol Fe2O3 (2 mol Fe/1 mol Fe2O3)( 56 g Fe/1 mol of Fe)= 36.06 g = 36.06/56= 0.643 mol
% yiled= actula yield x100/thereotical yield
= 10.3x100/36.06
= 28.5%
Combining 0.322 mol Fe,, with excess carbon produced 10.3 g Fe. Fe, 0, +3C — 2Fe...
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Combining 0.271 mol Fe2O3 with excess carbon produced 13.1 g Fe. Fe2O3+3C⟶2Fe+3CO What is the actual yield of iron in moles? actual yield: What is the theoretical yield of iron in moles? theoretical yield: What is the percent yield? percent yield:
Combining 0.296 mol of Fe_2O_3 with excess carbon produced 13.4 g of Fe. Fe_2O_3 + 3C rightarrow 2Fe + 3CO What is the actual yield of iron in moles? What was the theoretical yield of iron in moles? What was the percent yield?
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