Question

Question #d94d5

Answer 1

Here's what I got.

Explanation:

The trick here is to realize that the reaction involves white phosphorus, `"P"_4`, the most common allotrope of phosphorus, not atomic phosphorus, `"P"`, which only exists in the gaseous state.

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

`color(blue)(6)"Ca"_ ((s)) + "P"_ (4(s)) -> 2"Ca"_ 3"P"_ (2(s))`

Notice that the reaction consumes `color(blue)(6)` moles of calcium metal for every mole of white phosphorus and produces `2` moles of calcium phosphide, `"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 `"240.5 g"` of calcium for every `"123.9 g"` of white phosphorus that take part in the reaction.

The `color(blue)(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.