Gaseous butane CH3CH22CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 4.07 g of butane is mixed with 9.8 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to
2 significant digits.
This is a limiting reagent (LR) problem. This procedure will find the LR as well as finish the stoichiometry.
1. Write and balance the equation.
2a. Convert 5.07 g butane to mols. mols = grams/molar mass = ?
2b. Do the same and convert g O2 to mols.
3a. Using the coefficients in the balanced equation, convert mols butane to mols product (H2O).
3b. Do the same and convert mols O2 to mols H2O.
3c. It is likely that the two values for mols H2O will not be the same; the correct value in LR problems is ALWAYS the smaller one and the reagent responsible for this is the LR.
4. Using the smaller value of mols H2O, then grams = mols x molar mass = ?
Post your work if you get stuck.
To determine the maximum mass of water that could be produced, we need to determine the limiting reactant first. The limiting reactant is the reactant that is entirely consumed in the reaction and determines the maximum amount of product that can be formed.
We will compare the amount of water produced from both the butane and oxygen to determine the limiting reactant.
First, we need to calculate the number of moles of each reactant:
- Moles of butane = mass / molar mass = 4.07 g / 58.12 g/mol = 0.07 mol
- Moles of oxygen = mass / molar mass = 9.8 g / 32.00 g/mol = 0.31 mol
Next, we need to determine the stoichiometry of the reaction. From the balanced chemical equation:
C4H10 + 13/2 O2 → 4 CO2 + 5 H2O
We can see that 1 mole of butane (C4H10) reacts with 13/2 moles of oxygen (O2) to produce 5 moles of water (H2O).
Now, we compare the moles of water produced from both reactants.
- Moles of water from butane = moles of butane × (5 moles of water / 1 mole of butane) = 0.07 mol × (5 mol H2O / 1 mol butane) = 0.35 mol H2O
- Moles of water from oxygen = moles of oxygen × (5 moles of water / 13/2 moles of oxygen) = 0.31 mol × (5 mol H2O / 13/2 mol O2) = 0.12 mol H2O
Since the moles of water produced from butane (0.35 mol) is greater than the moles of water produced from oxygen (0.12 mol), oxygen is the limiting reactant.
Finally, we calculate the mass of water produced using the moles of water produced from oxygen:
Mass of water = moles of water from oxygen × molar mass of water = 0.12 mol × 18.02 g/mol = 2.16 g
Therefore, the maximum mass of water that could be produced is 2.16 g.
To calculate the maximum mass of water that could be produced in the given chemical reaction, we need to determine which reactant is limiting and calculate the amount of water produced using stoichiometry.
1. Calculate the molar mass of butane (C4H10):
Mass of C = 4 x atomic mass of C = 4.00 g/mol
Mass of H = 10 x atomic mass of H = 10.00 g/mol
Molar mass of butane (C4H10) = 4.00 g/mol + 10.00 g/mol = 58.12 g/mol
2. Calculate the molar mass of oxygen (O2):
Molar mass of O2 = 2 x atomic mass of O = 2.00 g/mol
3. Convert the given masses of butane and oxygen to moles:
Moles of butane = 4.07 g / 58.12 g/mol = 0.07 mol (rounded to two decimal places)
Moles of oxygen = 9.8 g / 2.00 g/mol = 4.9 mol (rounded to one decimal place)
4. Write and balance the chemical equation:
2 C4H10 + 13 O2 → 8 CO2 + 10 H2O
5. Determine the stoichiometry between butane and water:
According to the balanced equation, for every 2 moles of butane, 10 moles of water are formed.
6. Calculate the maximum moles of water that can be produced:
Based on the stoichiometry, the limiting reactant will determine the maximum moles of water produced. To do this, we need to find the limiting reactant.
The ratio of the moles of butane to the moles of oxygen is:
0.07 mol butane / 4.9 mol oxygen ≈ 0.014
The ratio indicates that there is an excess of oxygen compared to butane, which means the limiting reactant is butane.
Therefore, the maximum moles of water produced will be determined by the moles of butane: 0.07 mol.
7. Calculate the maximum mass of water produced:
Mass of water = Moles of water x Molar mass of water
Mass of water = 0.07 mol x (2 x atomic mass of H) = 0.07 mol x 2.02 g/mol
Mass of water = 0.142 g
So, the maximum mass of water that could be produced by the chemical reaction is 0.14 g.