The head of a match contains the chemical diphosphorus trisulfide.

Question

When this chemical undergoes friction it ignites and burns according to the equation given below:

 

P2S3(s) + 11/2 O2(g) -> P2O5(g) + 3SO2(g)

 

WHAT MASS OF P2S3 WOULD PRODUCE 310 ML OF SULFUR DIOXIDE GAS (SO2) AT PRESSURE OF 101.2 KPA AND A TEMPERATURE OF 37°C

Answer 1

Hello Tyler

 

First use the ideal gas law equation to calculate the # of moles of SO₂ gas.

 

PV = nRT    ;  rearrange to solve for n

 

n = PV / RT

 

R = 8.314 kPa L mol^-1 K^-1

 

V = 310 mL x (1 L / 1000 mL) = 0.31 L

 

T = 37 °C + 273K = 310 K

 

                 101.2 kPa x 0.31 L

n =  -----------------------------------------

        8.314 kPa L mol^-1 K^-1 x 310 K

 

    = 0.012 mol SO₂

 

Now use stoichiometry to convert moles of SO₂ to grams of P₂S₃.  From the balanced reaction, you can see that 1 mole of P₂S₃ reacts to form 3 moles of SO₂.  Molar mass of P₂S₃ = 158 g/mol

 

0.012 mol SO₂ x (1 mol P₂S₃ / 3 mol SO₂) x (158 g P₂S₃ / 1 mol P₂S₃)

 

= 0.63 g P₂S₃ will be needed