Question

In the following equation, how many moles of Al2O3 will be produced if 10 moles of O2 are reacted with excess Al? 4Al + 3O2 -> 2Al2O3HELPFUL INFO: molar mass of Al2O3=101.96g/mol, molar mass of O2=32g/mol, molar mass of Al=26.9g/mol

Answer 1

Since the balanced reaction is:

`4Al+3O_2\to2Al_2O_3`

The ratio os O₂ to Al₂O₃ will determine the number of moles produced.

We have excess of Al, so we can only look to O₂ and Al₂O₃.

Their coefficients are 3 for O₂ and 2 for Al₂O₃, so for every 3 moles of O₂, 2 moles of Al₂O₃ will be produced.

Using rules of three, we have:

Al₂O₃ --- O₂

x --- 10 mol

2 mol --- 3 mol

So we have the relation:

`\begin{gathered} (x)/(2mol)=(10)/(3) \\ x=2\cdot(10)/(3)mol \\ x=6.666\ldots mol\approx6.7mol \end{gathered}`

So, 10 mols of O₂ will produce approximately 6.7 mol of Al₂O₃.