In a butane lighter, 9.2g g of butane combines with 32.9g g of oxygen to form 27.8g g carbon dioxide and how many grams of water?
14.5g H2O
To find out the number of grams of water formed in the reaction of butane and oxygen, we need to first determine the balanced chemical equation for the reaction.
The balanced chemical equation for the combustion of butane (C₄H₁₀) is:
C₄H₁₀ + O₂ → CO₂ + H₂O
From the balanced equation, we can see that one mole of butane reacts with two moles of oxygen to produce one mole of carbon dioxide and one mole of water.
Let's calculate the number of moles of butane and oxygen based on the given masses:
Mass of butane (C₄H₁₀) = 9.2 g
Molar mass of butane (C₄H₁₀) = 58.12 g/mol
Number of moles of butane = Mass of butane / Molar mass of butane
Number of moles of butane = 9.2 g / 58.12 g/mol
Mass of oxygen (O₂) = 32.9 g
Molar mass of oxygen (O₂) = 32.00 g/mol
Number of moles of oxygen = Mass of oxygen / Molar mass of oxygen
Number of moles of oxygen = 32.9 g / 32.00 g/mol
Now, based on the balanced equation, we can determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the amount of product that can be formed.
From the balanced equation, we can see that the mole ratio between butane and oxygen is 1:2. Therefore, the limiting reactant is the one with fewer moles.
To calculate the moles of both reactants, divide their masses by their molar masses:
Number of moles of butane = 9.2 g / 58.12 g/mol ≈ 0.1585 mol
Number of moles of oxygen = 32.9 g / 32.00 g/mol ≈ 1.0281 mol
Since the mole ratio is 1:2, we can see that the butane is the limiting reactant because there are fewer moles of it.
Now, let's calculate the number of moles of carbon dioxide and water produced using the stoichiometry of the balanced equation.
From the balanced equation, we know that:
1 mole of butane produces 1 mole of carbon dioxide and 2 moles of water
Number of moles of carbon dioxide produced = Number of moles of butane = 0.1585 mol
Number of moles of water produced = 2 * Number of moles of butane = 2 * 0.1585 mol = 0.3170 mol
Finally, we can calculate the mass of water produced using the number of moles and the molar mass of water (H₂O):
Molar mass of water (H₂O) = 18.02 g/mol
Mass of water produced = Number of moles of water * Molar mass of water
Mass of water produced = 0.3170 mol * 18.02 g/mol
Therefore, the mass of water produced in the reaction is approximately 5.71 grams.
To find the mass of water produced in the reaction, we first need to figure out the balanced chemical equation for the combustion of butane (C4Hs):
C4H10 + O2 -> CO2 + H2O
From the balanced equation, we can see that for every mole of butane (C4H10) burned, we need 5 moles of oxygen (O2) to produce 4 moles of carbon dioxide (CO2) and 5 moles of water (H2O).
Now let's convert the given masses in grams into moles using the molar masses of each substance:
Molar mass of butane (C4H10) = 58.12 g/mol
Molar mass of oxygen (O2) = 32.00 g/mol
Molar mass of carbon dioxide (CO2) = 44.01 g/mol
Molar mass of water (H2O) = 18.02 g/mol
Mass of butane (C4H10) = 9.2 g
Mass of oxygen (O2) = 32.9 g
Mass of carbon dioxide (CO2) = 27.8 g
Now we can calculate the number of moles for each substance:
Number of moles of butane (C4H10) = 9.2 g / 58.12 g/mol ≈ 0.1586 mol
Number of moles of oxygen (O2) = 32.9 g / 32.00 g/mol ≈ 1.0281 mol
Number of moles of carbon dioxide (CO2) = 27.8 g / 44.01 g/mol ≈ 0.6310 mol
Since we know from the balanced equation that the molar ratio between carbon dioxide and water is 4:5, we can calculate the number of moles of water formed:
Number of moles of water (H2O) = (Number of moles of carbon dioxide) x (5 moles of water / 4 moles of carbon dioxide)
= 0.6310 mol x (5/4)
≈ 0.7888 mol
Finally, to find the mass of water produced, we multiply the number of moles by the molar mass of water:
Mass of water (H2O) = 0.7888 mol x 18.02 g/mol ≈ 14.2 g
Therefore, approximately 14.2 grams of water are produced in the combustion of butane with the given amounts of reactants.