1. Balanced Chemical Equation
The reaction you're describing is the reaction of calcium carbide (CaC₂) with water (H₂O) to produce acetylene (C₂H₂) and calcium hydroxide (Ca(OH)₂):
CaC₂ (s) + 2 H₂O (l) → C₂H₂ (g) + Ca(OH)₂ (aq)
2. Stoichiometry
* Molar Mass: Calculate the molar masses of the involved species:
* H₂O: 18.015 g/mol
* C₂H₂: 26.04 g/mol
* Mole Calculation: Convert the given mass of water (80 g) to moles:
* Moles of H₂O = (80 g) / (18.015 g/mol) = 4.44 mol
* Mole Ratio: Use the mole ratio from the balanced equation to find moles of C₂H₂:
* 2 mol H₂O : 1 mol C₂H₂
* Moles of C₂H₂ = (4.44 mol H₂O) * (1 mol C₂H₂ / 2 mol H₂O) = 2.22 mol C₂H₂
* Volume Calculation: We can use the Ideal Gas Law (PV = nRT) to convert moles of C₂H₂ to volume. We'll need to assume standard temperature and pressure (STP):
* STP: T = 273.15 K, P = 1 atm
* R (ideal gas constant) = 0.0821 L·atm/(mol·K)
* Volume of C₂H₂ = (nRT) / P = (2.22 mol * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm = 49.9 L
Therefore, approximately 49.9 liters of C₂H₂ should be produced from 80 grams of H₂O.
