When a mixture of 10.0 g of acetylene and 10.0 g of oxygen is ignited, a complete combustion of the reactants takes place. What is the limiting reagent of this reaction? How many grams of each of the reactants and products are present at the end of the reaction?
To determine the limiting reagent and the amount of each reactant and product, we need to compare the number of moles of acetylene and oxygen.
1. Calculate the number of moles for each reactant:
Number of moles of acetylene = mass of acetylene / molar mass of acetylene
Number of moles of oxygen = mass of oxygen / molar mass of oxygen
The molar mass of acetylene (C2H2) is 26.04 g/mol.
The molar mass of oxygen (O2) is 32.00 g/mol.
Number of moles of acetylene = 10.0 g / 26.04 g/mol = 0.384 mol
Number of moles of oxygen = 10.0 g / 32.00 g/mol = 0.313 mol
2. Determine the stoichiometric ratio of acetylene to oxygen based on the balanced equation. The balanced equation for the complete combustion of acetylene is:
2C2H2 + 5O2 -> 4CO2 + 2H2O
According to the equation, it takes 2 moles of acetylene to react with 5 moles of oxygen.
3. Compare the moles of acetylene and oxygen to identify the limiting reagent.
Since the stoichiometric ratio is 2:5, we need 2/5 times the number of moles of oxygen to react completely with acetylene.
Moles of oxygen needed = (2/5) * moles of acetylene = (2/5) * 0.384 mol = 0.1536 mol
Since the actual number of moles of oxygen (0.313 mol) is greater than the moles required (0.1536 mol), oxygen is in excess and acetylene is the limiting reagent.
4. Calculate the amount of each reactant and product at the end of the reaction:
Since acetylene is the limiting reagent, it will be completely consumed, and the amount of oxygen consumed will depend on the stoichiometric ratio.
Moles of oxygen consumed = moles of acetylene * (5/2) = 0.384 mol * (5/2) = 0.96 mol
Mass of acetylene remaining = moles of acetylene remaining * molar mass of acetylene
= 0 mol * 26.04 g/mol = 0 g
Mass of oxygen remaining = moles of oxygen remaining * molar mass of oxygen
= (moles of oxygen available - moles of oxygen consumed) * molar mass of oxygen
= (0.313 mol - 0.96 mol) * 32.00 g/mol = -21.984 g
Note: Negative mass means there is a deficiency in that reactant.
Mass of carbon dioxide formed = moles of carbon dioxide formed * molar mass of carbon dioxide
= moles of acetylene consumed * molar mass of carbon dioxide
= 0.384 mol * 44.01 g/mol = 16.89 g
Mass of water formed = moles of water formed * molar mass of water
= moles of acetylene consumed * molar mass of water
= 0.384 mol * 18.02 g/mol = 6.91 g
To summarize:
- The limiting reagent is acetylene.
- At the end of the reaction, there is 0 g of acetylene remaining.
- There is -21.984 g of oxygen remaining, indicating an excess.
- There are 16.89 g of carbon dioxide formed.
- There are 6.91 g of water formed.
To determine the limiting reagent and find the amounts of each reactant and product at the end of the reaction, you need to follow these steps:
1. Write the balanced chemical equation for the reaction:
C2H2 + O2 -> CO2 + H2O
2. Calculate the number of moles for each reactant using their respective molar masses:
Moles of C2H2 = Mass (g) / Molar mass (g/mol)
Moles of O2 = Mass (g) / Molar mass (g/mol)
Given:
Mass of C2H2 = 10.0 g
Molar mass of C2H2 = 26.04 g/mol
Moles of C2H2 = 10.0 g / 26.04 g/mol ≈ 0.384 mol
Mass of O2 = 10.0 g
Molar mass of O2 = 32.00 g/mol
Moles of O2 = 10.0 g / 32.00 g/mol ≈ 0.313 mol
3. Use the stoichiometry of the balanced chemical equation to determine the limiting reagent:
The balanced chemical equation tells us that the ratio of C2H2 to O2 is 1:1.
Since there is less O2 than C2H2, O2 is the limiting reagent.
4. Calculate the theoretical yield for the products using the limiting reagent:
In this reaction, 1 mole of C2H2 reacts with 1 mole of O2 to produce 1 mole of CO2 and 1 mole of H2O.
Therefore, the number of moles of CO2 and H2O produced is equal to the number of moles of O2.
Moles of CO2 = Moles of H2O = 0.313 mol
5. Calculate the masses of the products:
Mass of CO2 = Moles of CO2 * Molar mass of CO2
Mass of H2O = Moles of H2O * Molar mass of H2O
Given:
Molar mass of CO2 = 44.01 g/mol
Molar mass of H2O = 18.02 g/mol
Mass of CO2 = 0.313 mol * 44.01 g/mol ≈ 13.781 g
Mass of H2O = 0.313 mol * 18.02 g/mol ≈ 5.648 g
6. Determine the amount of each reactant and product remaining after the reaction:
Since O2 is the limiting reagent, all of the O2 is used up in the reaction. The amount of C2H2 remaining can be calculated by subtracting the moles of C2H2 used from the initial moles of C2H2.
Moles of C2H2 remaining = Initial moles of C2H2 - Moles of C2H2 used
= 0.384 mol - 0.313 mol ≈ 0.071 mol
Mass of C2H2 remaining = Moles of C2H2 remaining * Molar mass of C2H2
= 0.071 mol * 26.04 g/mol ≈ 1.854 g
Therefore, at the end of the reaction, there are approximately:
- 0.071 mol (1.854 g) of C2H2 remaining
- 0.000 mol (0 g) of O2 remaining (used up)
- 0.313 mol (13.781 g) of CO2 produced
- 0.313 mol (5.648 g) of H2O produced