Respuesta :
Answer:
a) 6.225 moles
b) 423.134grams
Explanations:
Given the balanced chemical equation as shown below:
[tex]4Al(s)+3O_2(g)\rightarrow2Al_2O_3[/tex]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:
[tex]\begin{gathered} mole\text{ of oxygen}=\frac{3moles\text{ of oxygen}}{4\cancel{moles\text{ of Al}}}\times8.3\cancel{moles\text{ of Al}} \\ mole\text{ of oxygen}=6.225moles \end{gathered}[/tex]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:
[tex]\begin{gathered} mole\text{ of Al}_2O_3=\frac{2moles\text{ of Al}_2O_3}{4\cancel{moles\text{ of Al}}}\times8.3\cancel{moles\text{ of Al}} \\ mole\text{ of Al}_2O_3=4.15moles \end{gathered}[/tex]Determine the mass of the product
[tex]\begin{gathered} Mass\text{ of Al}_2O_3=mole\times molar\text{ mass} \\ Mass\text{ of Al}_2O_3=4.15\times\frac{101.96g}{mol} \\ Mass\text{ of Al}_2O_3=423.134grams \end{gathered}[/tex]Hence the mass of product that will be formed from 8.3 moles of aluminum is 423.134grams
