Question

4 Al(s) + 3 O2(g) > 2 Al2O3(s)How many grams of Al2O3 can be formed from 32.19 g of Al and 50.22 g of O2?(Hint: You need to determine which one is the limiting reactant).

Answer 1

1) Chemical equation

`4Al_((s))+3O_(2(g))\rightarrow2Al_2O_(3(s))`

2) Convert grams to moles

Al: 32.19g

The molar mass of Al is 26.98 g/mol

`molAl=32.19gAl\cdot\frac{1molAl_{}}{26.98gAl_{}}=_{}1.19molAl_{}`

We have 1.19 mol Al.

O2: 50.22g

The molar mass of O2 is 31.999 g/mol

`molO_2=50.22gO_2\cdot(1molO_2)/(31.999gO_2)=1.57molO_2`

We have 1.57 mol O2.

3) Limiting reactant

How many moles of O2 do we need to use all of the Al?

The molar ratio between O2 and Al is 3 mol O2: 4 mol Al.

`molO_2=1.19molAl\cdot\frac{3molO_2}{4molAl_{}}_{}=0.8925molO_2`

We need 0.8925 mol O2 and we have 1.57 mol O2. We have enough O2.

O2 is the excess reactant.

How many moles of Al do we need to use all of the O2?

The molar ratio between O2 and Al is 3 mol O2: 4 mol Al.

`molAl=1.57molO_2\cdot\frac{4molAl_{}}{3molO_2}=2.09molAl_{}`

We need 2.09 mol Al and we have 1.19 mol Al. We do not have enough Al.

Al is the limiting reactant.

4) Moles of Al2O3 produced from 1.19 mol Al.

The molar ratio between Al and Al2O3 is 4 mol Al: 2 mol Al2O3.

`mol_{}Al_2O_3=1.19molAl\cdot(2molAl_2O_3)/(4molAl)_{}=0.595molAl_2O_3`

5) Convert moles to grams

The molar mass of Al2O3 is 101.96 g/mol

`gAl_2O_3=0.595molAl_2O_3\cdot(101.96gAl_2O_3)/(1molAl_2O_3)=60.67gAl_2O_3`

The mass of Al2O3 produced is 60.67 g.