Question #d94d5

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.