Calculate the mass of phosphorus pentasulfide P2S5, which can be obtained by fusing

Calculate the mass of phosphorus pentasulfide P2S5, which can be obtained by fusing sulfur with red phosphorus weighing 450 g, the mass fraction of impurities in which is 3%.

Given:

m (P) = 450 g

w% (impurities) = 3%

To find:

m (P2S5) -?

Decision:

2P + 5S = P2S5, – we solve the problem, relying on the composed reaction equation:

1) Find the mass of phosphorus without impurities:

if the proportion of impurities is 3%, then the proportion of pure phosphorus will be 97%,

then m (P) = 450 g * 0.97 = 436.5 g

2) Find the amount of phosphorus:

n (P) = m: M = 436.5 g: 31 g / mol = 14.08 mol

3) We compose a logical expression:

if 2 mol of P gives 1 mol of P2S5,

then 14.08 mol of P will give x mol of P2S5,

then x = 7.04 mol.

4) Find the mass of phosphorus sulfide formed during the reaction:

m (P2S5) = n * M = 7.04 mol * 222 g / mol = 1562.88 g.

Answer: m (P2S5) = 1562.88 g.