How much aluminum oxide is needed to prepare 617 g of aluminum by the balanced chemical reaction,2Al2O3(s)+3C(s)->4Al(s)+3CO2(g) if the reaction proceeds to 81.2 % yield?
First, let's calculate the molar mass of aluminum oxide (Al2O3):
2(Al) + 3(O) = 2(26.98 g/mol) + 3(16.00 g/mol) = 102.96 g/mol
Next, we need to determine the amount of aluminum oxide needed to produce 617 g of aluminum. Since the reaction is based on the molar ratio between aluminum and aluminum oxide, we need to convert the mass of aluminum to moles:
617 g / 26.98 g/mol = 22.88 moles of aluminum
Based on the balanced chemical equation, 2 moles of aluminum oxide are needed to produce 4 moles of aluminum. Therefore, we need:
22.88 moles of Al2O3 = 22.88 moles / 2 * 2 = 11.44 moles of Al2O3
Finally, we need to calculate the mass of aluminum oxide needed at 81.2% yield:
11.44 moles * 102.96 g/mol = 1177.48 g
However, since the reaction only proceeds to 81.2% yield, we need to adjust this calculation:
1177.48 g * 0.812 = 956.19 g
Therefore, 956.19 g of aluminum oxide is needed to produce 617 g of aluminum at 81.2% yield.