# How many moles of P4 are needed to react with 9 moles of Mg? 6 Mg + P4 → 2 Mg3P2

Jul 1, 2014

Let us look at the equation;

6Mg + ${P}_{4}$ .............>2 $M {g}_{3}$ ${P}_{2}$

according to this equation 1 mole of ${P}_{4}$ reacts with 6 moles of Mg. So my recipe is 1 mole of ${P}_{4}$ is required for every 6 moles of Mg.

1 mole of ${P}_{4}$ / 6 moles of Mg ......(a)

X moles of ${P}_{4}$ completely reacts with 9 moles of Mg

X moles of ${P}_{4}$ / 9 moles of Mg ......(b)

the two ratios are equal.

(a) = (b)

1 mole of ${P}_{4}$ / 6 moles of Mg =
X moles of ${P}_{4}$ / 9 moles of Mg

1. 1 = X .6

6X = 9

X = 9/6 = 3/2 = 1.5 mole