In a butane lighter, 9.8 g of butane combines with 35.1 g of oxygen to form 29.6 g carbon dioxide and how many grams of water?
To determine the number of grams of water formed, we'll need to use stoichiometry, which involves using the balanced chemical equation and the molar masses of the compounds involved.
The balanced chemical equation for the combustion of butane (C₄H₁₀) is as follows:
2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O
From the given information, we know that 9.8 g of butane (C₄H₁₀) combines with 35.1 g of oxygen (O₂) to form 29.6 g of carbon dioxide (CO₂).
To find the mass of water (H₂O) produced, we'll do the following calculations:
Step 1: Convert the masses of butane (C₄H₁₀) and oxygen (O₂) to moles using their molar masses.
The molar mass of butane (C₄H₁₀) = 4(12.01 g/mol) + 10(1.01 g/mol) = 58.12 g/mol
Number of moles of butane (C₄H₁₀) = 9.8 g / 58.12 g/mol
The molar mass of oxygen (O₂) = 2(16.00 g/mol) = 32.00 g/mol
Number of moles of oxygen (O₂) = 35.1 g / 32.00 g/mol
Step 2: Determine the limiting reactant by comparing the number of moles of each reactant.
Since the coefficients in the balanced equation show that 2 moles of butane react with 13 moles of oxygen, we can calculate the moles of butane needed for 35.1 g of oxygen by creating a proportion:
Number of moles of butane needed = (2 moles of butane / 13 moles of oxygen) * (Number of moles of oxygen)
If the number of moles of butane (C₄H₁₀) is greater than the number of moles of butane needed, then oxygen is the limiting reactant. Otherwise, butane is the limiting reactant.
Step 3: Calculate the moles of carbon dioxide (CO₂) produced from the limiting reactant.
From the balanced equation, we can see that 2 moles of butane react to give 8 moles of carbon dioxide. If butane is the limiting reactant, we can calculate the moles of carbon dioxide produced as follows:
Number of moles of carbon dioxide produced = (8 moles of carbon dioxide / 2 moles of butane) * (Number of moles of butane)
Step 4: Calculate the moles of water (H₂O) produced from the moles of carbon dioxide (CO₂) determined in Step 3.
From the balanced equation, we can see that 2 moles of butane react to give 10 moles of water. If butane is the limiting reactant, we can calculate the moles of water produced as follows:
Number of moles of water produced = (10 moles of water / 2 moles of butane) * (Number of moles of butane)
Step 5: Convert the moles of water to grams using the molar mass of water.
The molar mass of water (H₂O) = 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
Mass of water produced = Number of moles of water produced * Molar mass of water
By following these steps, you should be able to calculate the mass of water produced in the given reaction.
To determine the grams of water formed in the reaction, we need to use the balanced chemical equation for the combustion of butane:
C₄H₁₀ + 13/2 O₂ → 4 CO₂ + 5 H₂O
From the equation, we can see that for every mole of butane (C₄H₁₀) burned, we get 5 moles of water (H₂O) produced.
First, we need to calculate the number of moles of butane used:
Molar mass of butane (C₄H₁₀) = 12.01 g/mol (carbon) + 4 * 1.01 g/mol (hydrogen)
= 12.01 g/mol + 4.04 g/mol
= 16.05 g/mol
Moles of butane (C₄H₁₀) = mass / molar mass
= 9.8 g / 16.05 g/mol
≈ 0.61 mol
Now, using the stoichiometry of the balanced equation, we can calculate the moles of water produced:
Moles of water (H₂O) = 5 * moles of butane (C₄H₁₀)
= 5 * 0.61 mol
= 3.05 mol
Finally, we can determine the mass of water produced:
Molar mass of water (H₂O) = 2 * 1.01 g/mol (hydrogen) + 16.00 g/mol (oxygen)
= 2.02 g/mol + 16.00 g/mol
= 18.02 g/mol
Mass of water (H₂O) = moles * molar mass
= 3.05 mol * 18.02 g/mol
≈ 55.11 g
Therefore, approximately 55.11 grams of water would be produced in the butane lighter combustion process.