# Question d94d5

May 26, 2016

Here's what I got.

#### Explanation:

The trick here is to realize that the reaction involves white phosphorus, ${\text{P}}_{4}$, the most common allotrope of phosphorus, not atomic phosphorus, $\text{P}$, which only exists in the gaseous state.

This means that the balanced chemical equation for this reaction would look like this

$\textcolor{b l u e}{6} {\text{Ca"_ ((s)) + "P"_ (4(s)) -> 2"Ca"_ 3"P}}_{2 \left(s\right)}$

Notice that the reaction consumes $\textcolor{b l u e}{6}$ moles of calcium metal for every mole of white phosphorus and produces $2$ moles of calcium phosphide, ${\text{Ca"_3"P}}_{2}$.

Use the molar masses of the chemical species involved to convert the moles to grams. You will have

color(blue)(6)color(red)(cancel(color(black)("moles Ca"))) * "40.078 g"/(1color(red)(cancel(color(black)("mole Ca")))) = "240.5 g"

1 color(red)(cancel(color(black)("mole P"_4))) * "123.895 g"/(1color(red)(cancel(color(black)("mole P"_4)))) = "123.9 g"#

This means that the reaction consumes $\text{240.5 g}$ of calcium for every $\text{123.9 g}$ of white phosphorus that take part in the reaction.

The $\textcolor{b l u e}{6} : 1$ mole ratio is thus equivalent to a $240.5 : 123.9$ gram ratio. Use this ratio to determine how many grams of calcium are needed for a given amount of phosphorus.