Final answer:
When 80.0 grams of aluminum carbide and 80.0 grams of water are reacted, 17.78 grams of methane gas is produced, with water being the limiting reactant.
Step-by-step explanation:
The reaction between aluminum carbide (Al4C3) and water (H2O) produces methane gas (CH4) and aluminum hydroxide (Al(OH)3) as a precipitate. To determine the mass of CH4 produced when 80.0 grams of aluminum carbide is reacted with 80.0 grams of water, a stoichiometric calculation is required.
First, the molar masses of the reactants and products are calculated:
- Al4C3: 144 g/mol
- H2O: 18 g/mol
- CH4: 16 g/mol
- Al(OH)3: 78 g/mol
The balanced reaction is:
Al4C3 + 12H2O → 3CH4 + 4Al(OH)3
Next, find the moles of each reactant:
- Moles of Al4C3: 80.0 g / 144 g/mol = 0.556 mol
- Moles of H2O: 80.0 g / 18 g/mol = 4.444 mol
Since the reaction requires 12 moles of water for every mole of aluminum carbide, water is the limiting reactant. From this, we can calculate the amount of CH4 that can be produced:
- Moles of CH4: (4.444 mol H2O / 12 mol H2O/mol Al4C3) * 3 mol CH4/mol Al4C3 = 1.111 mol CH4
- Mass of CH4: 1.111 mol * 16 g/mol = 17.78 g
Therefore, 17.78 grams of methane gas will be produced when 80.0 grams of aluminum carbide is reacted with 80.0 grams of water, and water is the limiting reactant in this reaction.