How many moles of Mg3P2 can be produced from the reaction of .14 mol Mg(s) with .020 mol P4(s)
__Mg(s)+__P4(s)-->___Mg3P2

balance the equation first
__6Mg(s)+__P4(s)-->__2_Mg3P2
find the limiting reactant
convert moles of Mg into moles of Mg3P2
moles of P4 into moles of Mg3P4
the one which gives you less amount of the product is the limiting reactant and that will be the amount of the product formed also