Question

What is the theoretical yield of aluminum that can be produced by the reaction of 60.0 g of aluminum oxide with 30.0 g of carbon according to the following chemical equation? Al2O3 + 3C → 2Al + 3CO

Answer 1

Answer:

Theoretical yield = 31.8 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

For $Al_2O_3$

Mass of $Al_2O_3$  = 60.0 g

Molar mass of $Al_2O_3$  = 101.96128 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{60.0\ g}{101.96128\ g/mol}$

$Moles_{Al_2O_3}= 0.5885\ mol$

Given: For $C$

Given mass = 30.0 g

Molar mass of $C$  = 12.0107 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{30.0\ g}{12.0107\ g/mol}$

$Moles_{C}= 2.4978\ mol$

According to the given reaction:

$Al_2O_3+3C\rightarrow 2Al+3CO$

1 mole of aluminium oxide react with 3 moles of carbon

0.5885 mole of aluminium oxide react with $3\times 0.5885$ moles of carbon

Moles of carbon = 1.7655 moles

Available moles of carbon = 2.4978 moles

Limiting reagent is the one which is present in small amount. Thus, aluminium oxide is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of aluminium oxide on reaction forms 2 moles of aluminium.

0.5885 mole of aluminium oxide on reaction forms $2\times 0.5885$ moles of aluminium.

Moles of aluminium = 1.177 moles

Molar mass of aluminium = 26.981539 g/mol

Mass of sodium sulfate = Moles × Molar mass = 1.177 × 26.981539 g = 31.8 g

Theoretical yield = 31.8 g