a)What mass of water is produced from the complete combustion of 2.30×10−3 g of methane?
b)What mass of oxygen is needed for the complete combustion of 2.30×10−3 g of methane?
This is the unbalanced equation given:
CH4+O2->CO2+H2O
To solve these problems, we need to balance the chemical equation first:
CH4 + O2 -> CO2 + H2O
a) The equation shows that 1 mole of methane (CH4) produces 1 mole of water (H2O). We need to convert the given mass of methane into moles:
Molar mass of CH4:
C = 12.01 g/mol
H = 1.01 g/mol
Total molar mass of CH4 = 12.01 g/mol + (4 x 1.01 g/mol) = 16.05 g/mol
Given mass of CH4 = 2.30 × 10^−3 g
Number of moles of CH4 = (Given mass of CH4) / (Molar mass of CH4)
Number of moles of CH4 = (2.30 × 10^−3 g) / (16.05 g/mol)
Now we know how many moles of CH4 we have, and since the balanced equation tells us that 1 mole of CH4 produces 1 mole of H2O, the number of moles of water produced will be the same as the number of moles of CH4.
Number of moles of H2O = Number of moles of CH4
Number of moles of H2O = (2.30 × 10^−3 g) / (16.05 g/mol)
Finally, to find the mass of water produced, we can multiply the number of moles of H2O by its molar mass:
Molar mass of H2O:
H = 1.01 g/mol
O = 16.00 g/mol
Total molar mass of H2O = (2 x 1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
Mass of water produced = (Number of moles of H2O) x (Molar mass of H2O)
b) To determine the mass of oxygen needed for the complete combustion of methane, we first need to find the number of moles of methane (CH4) as calculated in part (a).
Since the balanced equation shows that 1 mole of CH4 reacts with 2 moles of O2, we can use the mole ratio to find the number of moles of O2 required:
Number of moles of O2 = 2 x (Number of moles of CH4)
Number of moles of O2 = 2 x [(2.30 × 10^−3 g) / (16.05 g/mol)]
To find the mass of oxygen required for the reaction, we multiply the number of moles of O2 by its molar mass:
Molar mass of O2: 16.00 g/mol + 16.00 g/mol = 32.00 g/mol
Mass of O2 required = (Number of moles of O2) x (Molar mass of O2)
By following these calculations, you can determine the mass of water produced and the mass of oxygen needed for the complete combustion of methane.
To determine the mass of water produced from the complete combustion of 2.30×10−3 g of methane, we need to use balanced equations and stoichiometry.
a) Balanced Equation:
CH4 + 2O2 -> CO2 + 2H2O
From the balanced equation, we can see that for every mole of methane (CH4) consumed, 2 moles of water (H2O) are produced.
1 mole of CH4 has a molar mass of 16.04 g, which is the sum of the atomic masses of carbon (12.01 g) and four hydrogen (1.01 g) atoms.
To calculate the number of moles of methane in 2.30×10−3 g, we use the formula: Moles = Mass / Molar mass
Moles of CH4 = 2.30×10−3 g / 16.04 g/mol = 1.43 × 10^-4 mol
Since the ratio of CH4 to H2O is 1:2, we can determine the number of moles of water produced:
Moles of H2O = 2 × Moles of CH4 = 2 × 1.43 × 10^-4 mol = 2.86 × 10^-4 mol
To find the mass of water produced, we use the formula: Mass = Moles × Molar mass
Mass of H2O = 2.86 × 10^-4 mol × 18.02 g/mol = 5.16 × 10^-3 g
Therefore, the mass of water produced from the complete combustion of 2.30×10−3 g of methane is approximately 5.16 × 10^-3 g.
b) To determine the mass of oxygen needed for the complete combustion of 2.30×10−3 g of methane, we use the stoichiometry provided by the balanced equation.
From the balanced equation:
1 mole of CH4 reacts with 2 moles of O2.
Using the same steps as before, we find that 1.43 × 10^-4 mol of CH4 reacts with:
Moles of O2 = 2 × Moles of CH4 = 2 × 1.43 × 10^-4 mol = 2.86 × 10^-4 mol
To calculate the mass of oxygen, we use the formula: Mass = Moles × Molar mass
Mass of O2 = 2.86 × 10^-4 mol × 32.00 g/mol = 9.15 × 10^-3 g
Therefore, the mass of oxygen needed for the complete combustion of 2.30×10−3 g of methane is approximately 9.15 × 10^-3 g.
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