1.78 mol. When 414 g of bismuth oxide react with excess carbon, 1.78 mol of bismuth form .
a) Start with the balanced chemical equation
2Bi2O3 + 3C → 4Bi + 3CO2
b) Convert grams of Bi2O3 to moles of Bi2O3
Moles of Bi2O3 = 414 g Bi2O3 x (1 mol Bi2O3/465.96 g Bi2O3) = 0.8885 mol Bi2O3
c) Use the molar ratio of Bi:Bi2O3 to convert moles of Bi2O3 to moles of Bi
Moles of Bi =0.8885 mol Bi2O3 x (4 mol Bi/2 mol Bi2O3) = 1.78 mol Bi