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notsponge [240]
3 years ago
11

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
Chemistry
1 answer:
Varvara68 [4.7K]3 years ago
5 0

Answer: 31.8 g

Explanation:

To calculate the moles :

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

\text{Moles of} Al_2O_3=\frac{60.0g}{102g/mol}=0.59moles

\text{Moles of} C=\frac{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=\frac{3}{1}\times 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 =\frac{2}{1}\times 0.59=1.18moles  of Al

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

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

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