What mass of water is produced from the complete combustion of 2.10×10−3g of methane?
Express your answer with the appropriate units.
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To solve this problem, we need to apply stoichiometry, which is the calculation of reactants and products in a chemical reaction.
The balanced chemical 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, 2 moles of water are produced.
1. Calculate the number of moles of methane (CH4):
To find the number of moles, we can use the formula: moles = mass / molar mass.
Given the mass of methane is 2.10×10−3g, we need to determine the molar mass of CH4. The molar mass of carbon (C) is 12.01 g/mol, and the molar mass of hydrogen (H) is 1.01 g/mol. Since methane has one carbon atom and four hydrogen atoms, the molar mass of CH4 is:
Molar mass of CH4 = (1 × molar mass of C) + (4 × molar mass of H)
= (1 × 12.01 g/mol) + (4 × 1.01 g/mol)
= 16.05 g/mol
Now, we can calculate the number of moles of methane using the formula:
moles of methane = mass of methane / molar mass of methane
= 2.10×10−3g / 16.05 g/mol
= 1.31×10−4 mol
2. Calculate the number of moles of water produced:
From the balanced chemical equation, we know that for every 1 mole of methane, 2 moles of water are produced. So, using stoichiometry, we can determine the number of moles of water produced:
moles of water = 2 × moles of methane
= 2 × 1.31×10−4 mol
= 2.62×10−4 mol
3. Calculate the mass of water produced:
Now, we can calculate the mass of water using the formula:
mass = moles × molar mass
The molar mass of water (H2O) is the sum of the molar masses of hydrogen (H) and oxygen (O):
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
mass of water = moles of water × molar mass of water
= 2.62×10−4 mol × 18.02 g/mol
= 4.72×10−3 g
Therefore, the mass of water produced from the complete combustion of 2.10×10−3 g of methane is 4.72×10−3 g.