Question

4 Al(s) + 302(g)2Al2O3(s)How many grams of Al2O3 can be formed from 99.20 g of Al and 30.22 g of O2?(Hint: You need to determine which one is the limiting reactant).

Answer 1

First, we have to convert each compound to moles, using the molar mass of each of them. The molar mass of Al is 27 g/mol and the molar mass of O2 is 32 g/mol (you can calculate the molar mass using the periodic table).

The conversion for 99.20 g of Al and 30.22 g of O2 would be:

`99.20\text{ g Al}\cdot\frac{1\text{ mol Al}}{27\text{ g Al}}=3.674\text{ moles Al.}`
`30.22gO_2\cdot(1molO_2)/(32gO_2)=0.9444molesO_2.`

Now, let's see how many moles of Al2O3 each of them produces. In the reaction, you can see that 4 moles of Al produce 2 moles of Al2O3:

`3.647\text{ moles Al}\cdot\frac{2molesAl_2O_3}{\text{4 moles Al}}=1.824\text{ moles }Al_2O_3.`

And 3 moles of O2 produce 2 moles of Al2O3:

`0.944molesO_2\cdot\frac{2\text{ moles }Al_2O_3}{3molesO_2}=0.6293\text{ moles }Al_2O_3.`

You can realize that we can just produce a minimum of 0.6293 moles of Al2O3 because there is an excess of Al, meaning that the limiting reactant is O2 and we take into account the number of moles of Al2O3 (0.6293 moles) produced by this compound.

The final step is to convert from 0.6293 moles of Al2O3 to grams using the molar mass of Al2O3 which is 102 g/mol:

`0.6293\text{ moles }Al_2O_3\cdot\frac{102\text{ g }Al_2O_3}{1\text{ mol }Al_2O_3}=64.19\text{ g }Al_2O_3.`

The answer would be that we're producing 64.19 g of Al2O3 from the reaction of 99.20 g of Al and 30.22 g of O2.