Question

_____Al (s)+ _____O2 (g)→ _____Al2O3 (s)a) How many moles of Oxygen are required to react with 8.3 moles of aluminum?b) How many grams of product will be formed from 8.3 moles of aluminum?

Answer 1

a) 6.225 moles

b) 423.134grams

Explanations:

Given the balanced chemical equation as shown below:

`4Al(s)+3O_2(g)\rightarrow2Al_2O_3`

a) Given the following parameters

Moles of aluminium = 8.3moles

According to stoichiometry, 4 moles of aluminium reacted with 3 moles of oxygen, the moles of oxygen are required to react with 8.3 moles of aluminum is expressed as:

`\begin{gathered} mole\text{ of oxygen}=\frac{3moles\text{ of oxygen}}{4\cancel{moles\text{ of Al}}}*8.3\cancel{moles\text{ of Al}} \\ mole\text{ of oxygen}=6.225moles \end{gathered}`

Hence the moles of oxygen required is 6.225moles

b) According to stoichiometry, 4 moles of Al produces 2 moles of aluminum oxide. The moles of aluminum oxide required will be:

`\begin{gathered} mole\text{ of Al}_2O_3=\frac{2moles\text{ of Al}_2O_3}{4\cancel{moles\text{ of Al}}}*8.3\cancel{moles\text{ of Al}} \\ mole\text{ of Al}_2O_3=4.15moles \end{gathered}`

Determine the mass of the product

`\begin{gathered} Mass\text{ of Al}_2O_3=mole* molar\text{ mass} \\ Mass\text{ of Al}_2O_3=4.15*(101.96g)/(mol) \\ Mass\text{ of Al}_2O_3=423.134grams \end{gathered}`

Hence the mass of product that will be formed from 8.3 moles of aluminum is 423.134grams