in the reaction 2al + cr2o3 give al2o3 + 2cr 49.8 g of al reacted with 200 g cr2o3 how much gram of reactant remains at the completion of the reaction
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Rewriting the equation :
2 Al + Cr₂O₃ —> Al₂O₃ + 2Cr
The mole ratio for the reactant is 2 : 1 meaning we need two moles of Aluminum to react with one mole of Cr₂O₃.
Moles for each reactant :
Cr₂O₃:
200 / 152 = 1.3158 moles.
Al :
49.8 / 26.98 = 2.6316
For the 1.3158 moles of Cr₂O₃ we need (2 × 1.3158) moles of Al.
Which equals to 2.6316moles.
We only have 1.8458 moles of Al which is less, meaning Al is the limiting reagent.
Thus Cr₂O₃ is in excess.
From the moles of Al, we need (1.8458/2) moles of Cr₂O₃ for complete reaction.
Which equals to 0.9229 moles.
The excess moles of Cr₂O₃ is :
1.3158 - 0.9229 = 0.3929 moles
From the molar mass we can get this mass :
152 × 0.3929 = 59.72 grams.
The excess reactant is thus Cr₂O₃ by mass of 59.72 grams.
2 Al + Cr₂O₃ —> Al₂O₃ + 2Cr
The mole ratio for the reactant is 2 : 1 meaning we need two moles of Aluminum to react with one mole of Cr₂O₃.
Moles for each reactant :
Cr₂O₃:
200 / 152 = 1.3158 moles.
Al :
49.8 / 26.98 = 2.6316
For the 1.3158 moles of Cr₂O₃ we need (2 × 1.3158) moles of Al.
Which equals to 2.6316moles.
We only have 1.8458 moles of Al which is less, meaning Al is the limiting reagent.
Thus Cr₂O₃ is in excess.
From the moles of Al, we need (1.8458/2) moles of Cr₂O₃ for complete reaction.
Which equals to 0.9229 moles.
The excess moles of Cr₂O₃ is :
1.3158 - 0.9229 = 0.3929 moles
From the molar mass we can get this mass :
152 × 0.3929 = 59.72 grams.
The excess reactant is thus Cr₂O₃ by mass of 59.72 grams.
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