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Anastaziya [24]

3 years ago

7

How many moles is 1g of Carbon?

1 answer:

7nadin3 [17]3 years ago

5 0

Answer:

The value of the mole is equal to the number of atoms in exactly 12 grams of pure carbon-12. 12.00 g C-12 = 1 mol C-12 atoms = 6.022 × 1023 atoms • The number of particles in 1 mole is called Avogadro's Number (6.0221421 x 1023).

Explanation:

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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

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<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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