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3 months ago

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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

1 answer:

Alekssandra [3K]3 months ago

3 0

Answer:

Theoretical yield = 31.8 g

Explanation:

The following formula is utilized to determine moles:

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

For $Al_2O_3$

Mass of $Al_2O_3$ = 60.0 g

Molar mass of $Al_2O_3$ = 101.96128 g/mol

The mole calculation formula is once again provided here:

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

Consequently,

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

$Moles_{Al_2O_3}= 0.5885\ mol$

Based on the provided reaction:

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

1 mole of aluminium oxide reacts with 3 moles of carbon

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

Moles of carbon = 1.7655 moles

Available moles of carbon = 2.4978 moles

The limiting reagent is defined as the substance present in the smallest quantity. Hence, aluminium oxide serves as the limiting reagent.

The yield of the product is dictated by the limiting reagent. Thus,

1 mole of aluminium oxide results in the formation of 2 moles of aluminium.

0.5885 mole of aluminium oxide results in the formation of $2\times 0.5885$ moles of aluminium.

Moles of aluminium = 1.177 moles

Molar mass of aluminium = 26.981539 g/mol

Mass of aluminium = Moles × Molar mass = 1.177 × 26.981539 g = 31.8 g

Theoretical yield = 31.8 g

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