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mojhsa [17]
3 years ago
11

When 125 grams of FeO react with 25.0 grams of Al, how many grams of Fe can be produced? FeO + Al → Fe + Al2O3 25.9 g Fe 38.7 g

Fe 50.9 g Fe 77.6 g Fe
Chemistry
1 answer:
Serga [27]3 years ago
4 0

<u>Answer:</u> The mass of iron produced will be 77.6 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For FeO:</u>

Given mass of FeO = 125 g

Molar mass of FeO = 71.8 g/mol

Putting values in equation 1, we get:

\text{Moles of FeO}=\frac{125g}{71.8g/mol}=1.74mol

  • <u>For aluminium:</u>

Given mass of aluminium = 25.0 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

\text{Moles of aluminium}=\frac{25.0g}{27g/mol}=0.93mol

The given chemical reaction follows:

3FeO+2Al\rightarrow 3Fe+Al_2O_3

By Stoichiometry of the reaction:

2 moles of aluminium metal reacts with 3 mole of FeO

So, 0.93 moles of aluminium metal will react with = \frac{3}{2}\times 0.93=1.395mol of FeO

As, given amount of FeO is more than the required amount. So, it is considered as an excess reagent.

Thus, aluminium metal is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of aluminium metal produces 3 mole of iron metal

So, 0.93 moles of aluminium metal will produce = \frac{3}{2}\times 0.93=1.395moles of iron metal

  • Now, calculating the mass of iron metal from equation 1, we get:

Molar mass of iron = 55.85 g/mol

Moles of iron = 1.395 moles

Putting values in equation 1, we get:

1.395mol=\frac{\text{Mass of iron}}{55.85g/mol}\\\\\text{Mass of iron}=(1.395mol\times 55.85g/mol)=77.6g

Hence, the mass of iron produced will be 77.6 grams

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Answer:

Here's what I get.

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The microdistillation procedure will vary, depending on the specific apparatus you are using, but here is a typical procedure.

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