Extra Stoichiometry Practice
1. How many grams of aluminum oxide can be formed by the reaction of 38.8 g of aluminum with
oxygen?
4 Al (8) + 3 O2(g) →2 A12O(s)

50. 4g

Explanation:

First calculate number of moles of aluminium in 38.8g

Moles = 38.8g/ 26.982mol/g

= 1.44mol

By looking at the balance equation you can see that 4 moles of aluminium produce 2 moles of aluminium oxide.

4 = 2

1.4 = x

Find the value of x

x= (1.4×2)/4= 0.72 mol

0.72 moles of aluminium oxide are produced from 38.8g of aluminium

Now find the mass of aluminium produced.

Mass = moles × molar mass

= 0.72mol × 69.93 mol/g

= 50.4g

the law of conservation of mass states that in a chemical reaction, the total mass of reactants is equal to the total mass of products. because atoms are only rearranged in a chemical reaction, there must be the same number of sodium atoms and chlorine atoms in both the reactants and products

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