Determine the mass of aluminum (Al) produced in the reaction if 30.00 grams of magnesium (Mg) was consumed with the equation given; 2 AlPO4(aq) + 3 Mg(s) = 2 Al(s) + 1 Mg3(PO4)2(aq)

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

AlPO₄: 2 moles
Mg: 3 moles
Al: 2 moles
Mg₃(PO₄)₂: 1 mole

Being the molar mass of each compound:

AlPO₄: 122 g/mole
Mg: 24.31 g/mole
Al: 27 g/mole
Mg₃(PO₄)₂: 262.93 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

AlPO₄: 2 moles* 122 g/mole= 244 grams
Mg: 3 moles* 24.31 g/mole= 72.93 grams
Al: 2 moles* 27 g/mole= 54 grams
Mg₃(PO₄)₂: 1 mole* 262.93 g/mole= 262.93 grams

You can apply the following rule of three: if by stoichiometry 72.93 grams of Mg produces 54 grams of Al, 30 grams of Mg produces how much mass of Al?

[tex]mass of Al=\frac{30 grams of Mg*54 grams of Al}{72.93 grams of Mg}[/tex]

mass of Al= 22.21 grams

22.21 grams of aluminum (Al) was produced in the reaction if 30.00 grams of magnesium (Mg) was consumed.