Gaseous butane
CH3CH22CH3
will react with gaseous oxygen
O2
to produce gaseous carbon dioxide
CO2
and gaseous water
H2O
. Suppose 36.0 g of butane is mixed with 41. g of oxygen. Calculate the minimum mass of butane that could be left over by the chemical reaction.
From the question I assume we know that O2 is the limiting reagent and butane is the excess reagent.
mols butane = grams/molar mass = ?
mols O2 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols O2 to mols butane. That will give you the mols butane used by the O2. Subtract initial mols butane - mols butane used by the O2 reaction = mols butane left. Then convert mols butane to grams by g butane = mols butane x molar mass butane.
To calculate the minimum mass of butane that could be left over after the reaction, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
First, we need to calculate the moles of each reactant using their molar masses:
Molar mass of butane (C4H10) = (12.01 g/mol × 4) + (1.01 g/mol × 10) = 58.12 g/mol
Molar mass of oxygen (O2) = 16.00 g/mol × 2 = 32.00 g/mol
Number of moles of butane = mass of butane / molar mass of butane = 36.0 g / 58.12 g/mol
Number of moles of oxygen = mass of oxygen / molar mass of oxygen = 41.0 g / 32.00 g/mol
Next, we need to determine the mole ratio between butane and oxygen in the balanced chemical equation. From the equation:
2C4H10 + 13O2 -> 8CO2 + 10H2O
We can see that the mole ratio between butane and oxygen is 2:13.
Now, we compare the moles of butane and oxygen using the mole ratio to determine the limiting reactant.
Number of moles of butane required = (number of moles of oxygen) × (2 moles of butane / 13 moles of oxygen)
If the number of moles of butane required is greater than the number of moles of butane given, then oxygen is the limiting reactant. Otherwise, butane is the limiting reactant.
Now we can calculate the minimum mass of butane that could be left over.
If oxygen is the limiting reactant:
- Determine the maximum moles of butane that can react by using the mole ratio.
- Calculate the mass of the remaining butane by subtracting the mass of reacted butane from the initial mass.
If butane is the limiting reactant:
- All the butane will react, and there will be no remaining butane.
I'll calculate the values for you.
To determine the minimum mass of butane that could be left over after the chemical reaction, we need to first calculate the amount of butane and oxygen consumed in the reaction.
1. Calculate the molar mass of butane (C4H10):
- Carbon (C): 4 x 12.01 g/mol = 48.04 g/mol
- Hydrogen (H): 10 x 1.01 g/mol = 10.10 g/mol
Total molar mass of butane = 48.04 g/mol + 10.10 g/mol = 58.14 g/mol
2. Calculate the molar mass of oxygen (O2):
- Oxygen (O): 2 x 16.00 g/mol = 32.00 g/mol
3. Calculate the moles of butane:
Moles of butane = mass of butane / molar mass of butane
Moles of butane = 36.0 g / 58.14 g/mol ≈ 0.619 mol
4. Calculate the moles of oxygen:
Moles of oxygen = mass of oxygen / molar mass of oxygen
Moles of oxygen = 41.0 g / 32.00 g/mol ≈ 1.281 mol
5. Determine the limiting reactant (the reactant that is completely consumed):
The balanced chemical equation for the reaction is:
2 C4H10 + 13 O2 → 8CO2 + 10H2O
From the balanced equation, we can see that 2 moles of butane react with 13 moles of oxygen. Therefore, the oxygen is limiting the reaction since it requires more moles.
6. Calculate the moles of butane required (based on the stoichiometry of the reaction):
Moles of butane required = (moles of oxygen) x (2 moles of butane / 13 moles of oxygen)
Moles of butane required = 1.281 mol x (2 mol / 13 mol) ≈ 0.197 mol
7. Calculate the mass of butane required:
Mass of butane required = (moles of butane required) x (molar mass of butane)
Mass of butane required = 0.197 mol x 58.14 g/mol ≈ 11.45 g
8. Calculate the mass of butane left over:
Mass of butane left over = initial mass of butane - mass of butane required
Mass of butane left over = 36.0 g - 11.45 g ≈ 24.55 g
Therefore, the minimum mass of butane that could be left over by the chemical reaction is approximately 24.55 grams.