Gas Stoichiometry

Question

If acetylene is burned in a combustion reaction,

a) What is the number of grams in acetylene would be needed to produce 63 liters of CO2 at the temperature of 1000 C?

b) How many grams (in water) would this produce?

Answer 1

2C₂H₂(g) + 5O₂ (g) ⟶ 4CO₂ (g) + 2H₂O(g)

1.96 g/L is the density of CO2

63 L CO2 x 1.96 g/ 1 L =123.48 g CO2

This step is used to get CO2 from L to grams. Note the L drop out when L is divided by L.

Now we need to get from CO2 in grams to moles, the central unit.

We will use the molecular weight of CO2: 44.01 g/mol. We need grams in the denominator to get grams to cancel out, so we will multiply by 1 mol/44.01 grams

123.48 g CO2 x 1 mol/44.01 g= 2.81 mol CO2

Now we will convert CO2 to acetylene (C2H2); for 4 moles of CO2, we need 2 moles of C2H2. Divide my mol CO2 to get moles CO2 to cancel out.

2.81 mol CO2 x 2 mol C2H2 / 4 mol CO2 = 1.41 mol C2H2

Last step, we need grams, so we will multiply by C2H2 molecular mass which is 26.038 g/mol.

1.41 mol C2H2 x 26.038 g / 1 mol= 36.7 g C2H2

For part b, we will do the same steps and convert it to H2O

36.7 g C2H2 x 1 mol C2H2 / 26.038 g = 1.41 mol C2H2

1.41 mol C2H2 x 2 mol H2O / 2 mol C2H2 = 1.41 mol H2O

1.41 mol H2O x 18.02 g / 1 mol H2O = 25.4 g H2O