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? ALLO, + 3C SH 2Al + 3CO

a. 31.8 g
b. 30g
c. 101.2 g
d. 45 g
e. 7.9 g

Answer 1

Answer: 31.8 g

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} Al_2O_3=(60.0g)/(102g/mol)=0.59moles`

`\text{Moles of} C=(30.0g)/(12g/mol)=2.5moles`

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

According to stoichiometry :

1 mole of
`Al_2O_3` require 3 moles of
`C`

Thus 0.59 moles of
`Al_2O_3` will require=
`(3)/(1)* 0.59=1.77moles` of
`C`

Thus
`Al_2O_3` is the limiting reagent as it limits the formation of product and
`C` is the excess reagent as it is present in more amount than required.

As 1 mole of
`Al_2O_3` give = 2 moles of
`Al`

Thus 0.59 moles of
`Al_2O_3` give =
`(2)/(1)* 0.59=1.18moles` of
`Al`

Mass of
`Al=moles* {\text {Molar mass}}=1.18moles* 27g/mol=31.8g`

Thus 31.8 g of
`Al` will be produced from the given masses of both reactants.