Question

The compound P₄S₃ is used in matches. It reacts with oxygen to produce P₄O₁₀ and SO₂. The unbalanced chemical equation is shown below:

P₄S₃(s) + O₂(g) --> P₄O₁₀(s) + SO₂(g)

What mass of SO₂ is produced from the combustion of 0.401 g P₄S₃?

Answer 1

Answer – Given, mass of P₄S₃ = 0.401 g

Reaction –

P₄S₃(s) + 8 O₂(g) ----> P₄O₁₀(s) + 3 SO₂(g)

Calculation of the mole of the 0.401 g of P₄S₃(s)

Moles of P₄S₃(s) = 0.401 g / 220.094 g.mol^-1

                            = 0.00182 moles

Now from the above balanced equation

1 moles of P₄S₃(s) = 3 mole of SO₂(g)

So, 0.00182 moles of P₄S₃(s) = /

= 0.00547 moles of SO₂(g)

Now we need to convert this moles of SO₂(g) to its mass

Mass of SO₂(g) = moles of SO₂(g) * molar mass of SO₂(g)

                          = 0.00547 moles * 64.064 g/mol

                          = 0.350 g of SO₂(g)

So, 0.350 g of mass of SO₂ is produced from the combustion of 0.401 g P₄S₃