Calculate the mass of the substance formed by the interaction of 2.4 g of calcium with an excess of phosphorus.

Given:

m (Ca) = 2.4 g;

To find:

m (Ca3P2) -?

Decision:

1) We compose the reaction equation typical for the condition of the given problem:

3Ca + 2P = Ca3P2;

2) Find the amount of calcium contained in 2.4 grams of metal:

n (Ca) = m: M = 2.4 g: 40 g / mol = 0.06 mol;

3) We compose logical equality:

if 3 mol of Ca gives 1 mol of Ca3P2,

then 0.06 mol of Ca will give x mol of Ca3P2,

then x = 0.02 mol.

4) Find the mass of calcium phosphide formed as a result of the reaction:

m (Ca3P2) = n * M = 0.02 mol * 182 g / mol = 3.64 g;

Answer: m (Ca3P2) = 3.64 g.