Chemistry, asked by harmeet5081, 1 year ago

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

Answers

Answered by danielochich
21
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.

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