Question

Consider the reaction: 4Al + 3MnO2 è 3Mn + 2Al2O3. if 22 grams of Al react with 13 grams of manganese (IV) oxide, how much excess reactant remains aRer the reac/on has run to completion?

Answer 1

Answer: According to the given informations of the question we know that 4 mol of Al will react with 3 mole of manganese(IV) oxide .

Also we can write it as , 108 g of Al wii react with = 261 g of MnO2

Hence , 22 g of Al will react with = 261 * 22 / 108 g of MnO2

= 53.16 g of MnO2

and in question we have only 13 gram of MNO2 hence the limiting reagent is MnO2 and Al wii reamins in reagent after complition of reaction

Now , 261 g MNO2 required = 108 g Al

Then , 13 Gram will required = 108 * 13 / 261 g of Al = 5.379 g

hence the amount of Al will remain in after completion of reaction is 16.62 g

This is all about the given question . Thank you . :)