2C2H6(g)+7O2(g)--> 4Co2 (g)+6H2O(l)
A) if 3.00 moles of C2H6 and 9.00 moles of O2 are introducedinto an empty container at 32.0 degree C and then ignited tointiate the above reaction calculate the mass of water that isproduced.
B) if the container volume is 675 what mass of water is in thevapour phase?
To solve both parts A and B of this question, we need to use stoichiometry, which is the calculation of the relative quantities of reactants and products in a chemical reaction. We will also need the molar masses of the compounds involved.
A) To calculate the mass of water produced:
Step 1: Balance the equation by ensuring that the number of atoms of each element is the same on both sides of the equation:
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)
Step 2: Determine the molar masses of the compounds involved:
- Molar mass of C2H6 = (2 * Molar mass of C) + (6 * Molar mass of H) = (2 * 12.01 g/mol) + (6 * 1.01 g/mol) = 30.07 g/mol
- Molar mass of O2 = (2 * Molar mass of O) = (2 * 16.00 g/mol) = 32.00 g/mol
- Molar mass of CO2 = (1 * Molar mass of C) + (2 * Molar mass of O) = (1 * 12.01 g/mol) + (2 * 16.00 g/mol) = 44.01 g/mol
- Molar mass of H2O = (2 * Molar mass of H) + (1 * Molar mass of O) = (2 * 1.01 g/mol) + (1 * 16.00 g/mol) = 18.02 g/mol
Step 3: Use stoichiometry to determine the moles of water produced:
- From the balanced equation, the stoichiometric ratio between C2H6 and H2O is 2:6. This means that for every 2 moles of C2H6, 6 moles of H2O are produced.
- Moles of C2H6 = 3.00 mol (given)
- Moles of H2O = (6/2) * Moles of C2H6 = (6/2) * 3.00 mol = 9.00 mol
Step 4: Calculate the mass of water produced:
- Mass of H2O = Moles of H2O * Molar mass of H2O = 9.00 mol * 18.02 g/mol = 162.18 g
Therefore, the mass of water produced is 162.18 g.
B) To calculate the mass of water in the vapor phase:
Step 1: Determine the moles of water in the vapor phase:
- From the balanced equation, we know that the stoichiometric ratio between CO2 and H2O is 4:6. This means that for every 4 moles of CO2, 6 moles of H2O are produced.
- Moles of CO2 = 4/2 * Moles of C2H6 = 4/2 * 3.00 mol = 6.00 mol
- Moles of H2O = (6/4) * Moles of CO2 = (6/4) * 6.00 mol = 9.00 mol
Step 2: Calculate the volume of water in the vapor phase using the ideal gas law:
- Volume of water in the vapor phase = Moles of H2O * Gas constant * Temperature / Pressure
- Gas constant = 0.0821 L·atm/(mol·K) (atm is the unit of pressure in this case)
- Temperature = 32.0 °C = 32.0 + 273.15 = 305.15 K (converted to Kelvin)
- Pressure is not given, so it cannot be calculated. Without pressure, we cannot determine the volume.
Therefore, without the pressure value, we cannot calculate the mass of water in the vapor phase.