In a butane lighter, 9.5 of butane combines with 34.0 of oxygen to form 28.7 carbon dioxide and how many grams of water?
9.5 g Butane + 34.0 g Oxygen = 43.5
Then 43.5 - 28.7 g CO2 = 14.8
That is the answer: 14.8 g
9.5 what
34.0 what
28.7 what
To find the mass of water formed in the reaction, we can use the balanced chemical equation for the combustion of butane:
2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O
From the equation, we can see that for every 2 moles of butane (C4H10), 10 moles of water (H2O) are formed.
First, let's convert the given mass of butane and oxygen to moles:
Mass of butane = 9.5 g
Molar mass of butane (C4H10) = 58.12 g/mol
Moles of butane = Mass / Molar mass = 9.5 g / 58.12 g/mol ≈ 0.1635 mol
Mass of oxygen = 34.0 g
Molar mass of oxygen (O2) = 32.00 g/mol
Moles of oxygen = Mass / Molar mass = 34.0 g / 32.00 g/mol ≈ 1.0625 mol
According to the chemical equation, the reactant ratio between butane and water is 2:10.
Moles of water = Moles of butane × (10 moles of water / 2 moles of butane) = 0.1635 mol × 5 = 0.8175 mol
Now, let's calculate the mass of water formed:
Molar mass of water (H2O) = 18.02 g/mol
Mass of water = Moles × Molar mass = 0.8175 mol × 18.02 g/mol ≈ 14.75285 g ≈ 14.8 g
Therefore, approximately 14.8 grams of water are formed in the combustion of 9.5 grams of butane with 34.0 grams of oxygen.
To find the grams of water formed, we need to use the balanced chemical equation for the combustion of butane:
C4H10 + O2 → CO2 + H2O
From the equation, we see that for every 1 mole of butane (C4H10) burnt, 2 moles of water (H2O) are formed.
Step 1: Convert given masses to moles
Moles of butane (C4H10):
Given mass = 9.5 g
Molar mass of butane = 58.12 g/mol
Moles of butane = Given mass / Molar mass
Moles of butane = 9.5 g / 58.12 g/mol
Moles of oxygen (O2):
Given mass = 34.0 g
Molar mass of oxygen = 32.00 g/mol
Moles of oxygen = Given mass / Molar mass
Moles of oxygen = 34.0 g / 32.00 g/mol
Step 2: Determine the limiting reactant
The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed. To find the limiting reactant, we compare the moles of reactants and their stoichiometric ratio.
According to the balanced equation, 1 mole of butane reacts with 13/2 moles of oxygen to form 7/2 moles of water. Therefore, we need to compare the moles of oxygen and butane in their ratio (13/2 : 1) to determine which is limiting.
Ratio of moles of butane to moles of oxygen = (9.5 g butane / 58.12 g/mol) : (34.0 g oxygen / 32.00 g/mol)
Step 3: Calculate the moles of water formed
Since the limiting reactant is determined by the ratio of moles, we use the moles of oxygen in the ratio (13/2 : 1) to calculate the moles of water formed.
Moles of water = Moles of oxygen × (7/2 moles of water / (13/2 moles of oxygen))
Step 4: Convert moles of water to grams
To find the mass of water formed, we use the equation:
Mass = Moles × Molar mass
Substitute the calculated moles of water into the equation to find the mass of water formed. The molar mass of water (H2O) is 18.015 g/mol.
Mass of water = Moles of water × Molar mass of water
Finally, you will have the grams of water formed.