What mass of water is produced from the complete combustion of 2.60×10−3 of methane?
Express your answer with the appropriate units.
Same example.
http://www.jiskha.com/science/chemistry/stoichiometry.html
What is the the PH change in the following reaction...
H2CO3 + 2KOH -> K2CO3 + 2H2O
To determine the mass of water produced from the complete combustion of methane, we need to use the balanced chemical equation for the combustion reaction.
The balanced equation for the combustion of methane (CH4) is:
CH4 + 2O2 → CO2 + 2H2O
From the equation, we can see that for every 1 mole of methane burned, we get 2 moles of water produced.
Given that the amount of methane is 2.60 × 10^(-3) moles, we can calculate the moles of water produced using the ratio of moles between methane and water.
Moles of water = (2.60 × 10^(-3) mol CH4) × (2 mol H2O / 1 mol CH4)
Moles of water = 5.20 × 10^(-3) mol H2O
Now, we need to convert the moles of water to mass:
Molar mass of water (H2O) = (2 × atomic mass of hydrogen) + atomic mass of oxygen
= (2 × 1.008 g/mol) + 15.999 g/mol
= 18.015 g/mol
Mass of water = (5.20 × 10^(-3) mol H2O) × (18.015 g/mol)
Mass of water = 0.0936 g
Therefore, the mass of water produced from the complete combustion of 2.60 × 10^(-3) moles of methane is 0.0936 grams.
To find the mass of water produced from the complete combustion of methane, you need to use the balanced chemical equation of the reaction.
The balanced equation for the combustion of methane is:
CH4 + 2O2 → CO2 + 2H2O
From this equation, we can see that for every 1 mole of CH4 burned, we get 2 moles of H2O formed.
To solve the problem, follow these steps:
1. Convert the given mass of methane to moles.
Given mass of methane = 2.60 × 10^−3 g
Molar mass of methane (CH4) = 12.01 g/mol + 1.01 g/mol × 4 = 16.04 g/mol
Moles of CH4 = (given mass of methane) / (molar mass of methane)
= (2.60 × 10^−3 g) / (16.04 g/mol)
2. Use the mole ratio from the balanced equation to find the moles of water produced.
Moles of H2O = (moles of CH4) × (2 moles H2O / 1 mole CH4)
3. Convert moles of water to mass.
Molar mass of water (H2O) = 2.02 g/mol
Mass of water = (moles of H2O) × (molar mass of H2O)
= (moles of H2O) × (2.02 g/mol)
Now, plug in the values and calculate:
Moles of CH4 = (2.60 × 10^−3 g) / (16.04 g/mol) ≈ 1.621 × 10^−4 mol
Moles of H2O = (1.621 × 10^−4 mol) × (2 mol H2O / 1 mol CH4) ≈ 3.242 × 10^−4 mol
Mass of water = (3.242 × 10^−4 mol) × (2.02 g/mol) ≈ 6.551 × 10^−4 g
Therefore, the mass of water produced from the complete combustion of 2.60 × 10^−3 g of methane is approximately 6.551 × 10^−4 g.