A certain lamp produces 10.0 L of CO₂(g) at STP. How many grams of calcium carbide must have reacted in the lamp?

Answer 1

To address this question, we need to first write the correctly balanced equation for the reaction that is taking place to produce the CO₂(g). This is known as a carbide lamp, or an acetylene gas lamp.

Calcium carbide = CaC₂

The reaction is:

CaC₂ + 2H₂O ==> C₂H₂ + Ca(OH)₂ where C₂H₂ is acetylene

2C₂H₂ + 5 O₂ ==> 4CO₂ + 2H₂O

So we see that for every 1 mole of CaC₂, we obtain 4 moles of CO₂.

We can now use this relationship to answer the question.

At STP one mole of an ideal gas = 22.4 L

10.0 L CO₂ x 1 mol / 22.4 L = 0.446 moles CO₂

moles CaC₂ used = 0.446 mols CO₂ x 1 mol CaC₂ / 4 mols CO₂ = 0.112 mols CaC₂ used

molar mass CaC₂ = 64.1 g/mol

grams CaC₂ reacted = 0.112 mols CaC₂ x 64.1 g / mol = 7.15 g CaC₂ (3 sig. figs.)