Fe2o3 + 2al -> al2o3 + 2fe

calculate the mass of iron metal (in grams) that can be prepared from 150 grams of aluminum and 250 grams of iron(iii) oxide.

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Answer : The mass of iron metal needed are, 87.4 grams

Solution : Given,

Mass of Al = 150 g

Mass of = 250 g

Molar mass of Al = 27 g/mole

Molar mass of = 160 g/mole

Molar mass of Fe = 56 g/mole

First we have to calculate the moles of Al and .

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

From the balanced reaction we conclude that

As, 2 mole of react with 1 mole of

So, 1.56 moles of react with moles of

From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of

From the reaction, we conclude that

As, 2 mole of react to give 2 mole of

So, 1.56 mole of react to give 1.56 mole of

Now we have to calculate the mass of

Thus, the mass of iron metal needed are, 87.4 grams

h let me think