In a butane lighter, 9.2 g of butane combines with 32.9 g of oxygen to form 27.8 g carbon dioxide and how many grams of water?
2C4H10 + 13O2 ==> 8CO2 + 10H2O
9.2..............32.9............27.8......?
I can't read the mind of the problem's author but I suspect this is not a limiting reagent problem. If not it can be done another way although what oobleck has shown will ALWAYS work.
I think the problem is getting at the Law of Conservation of Mass, therefore
input grams = output grams
(9.2 + 32.9) = 27.9 + mass H2O. Solve for mass H2O
how many moles of butane and oxygen are present?
write the equation
which one will get used up first?
the equation will tell you how many moles of water to expect
convert that to grams
To find the number of grams of water produced, we need to use the concept of balancing chemical equations and the law of conservation of mass.
Step 1: Write the balanced chemical equation for the combustion of butane (C4H10) with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O).
C4H10 + O2 -> CO2 + H2O
Step 2: Determine the molar masses of the compounds involved in the reaction:
- Butane (C4H10) has a molar mass of 58.12 g/mol
- Oxygen (O2) has a molar mass of 32.00 g/mol
- Carbon dioxide (CO2) has a molar mass of 44.01 g/mol
- Water (H2O) has a molar mass of 18.02 g/mol
Step 3: Calculate the number of moles of butane (C4H10) and oxygen (O2) used in the reaction using their molar masses and given masses:
- Moles of butane = mass of butane / molar mass of butane = 9.2 g / 58.12 g/mol = 0.158 moles
- Moles of oxygen = mass of oxygen / molar mass of oxygen = 32.9 g / 32.00 g/mol = 1.028 moles
Step 4: Use the balanced chemical equation to determine the stoichiometry of the reaction. The mole ratio between butane, oxygen, carbon dioxide, and water is 1:13:8:10.
- Moles of carbon dioxide = 0.158 moles butane x (8 moles CO2 / 1 mole butane) = 1.264 moles
- Moles of water = 0.158 moles butane x (10 moles H2O / 1 mole butane) = 1.58 moles
Step 5: Calculate the mass of water produced using its molar mass and the number of moles determined:
- Mass of water = moles of water x molar mass of water = 1.58 moles x 18.02 g/mol = 28.476 g
So, approximately 28.476 grams of water will be produced.