39.3 grams of c6h6 are allowed to react with 108.3 grams of o2. How much co2 will be produced by this reaction?
To find out how much CO2 will be produced by the reaction between C6H6 (Benzene) and O2 (oxygen), we need to balance the chemical equation and calculate the stoichiometry.
The balanced chemical equation for the combustion of benzene is:
C6H6 + 15O2 -> 6CO2 + 3H2O
Given:
Mass of C6H6 = 39.3 grams
Mass of O2 = 108.3 grams
To calculate the amount of CO2 produced, we'll follow these steps:
Step 1: Calculate the number of moles of C6H6 and O2 using their molar masses.
Molar mass of C6H6: (6 x 12.01 g/mol) + (6 x 1.01 g/mol) = 78.11 g/mol
Number of moles of C6H6 = Mass / Molar mass = 39.3 g / 78.11 g/mol
Molar mass of O2: 2 x 16.00 g/mol = 32.00 g/mol
Number of moles of O2 = Mass / Molar mass = 108.3 g / 32.00 g/mol
Step 2: Determine the limiting reactant. The limiting reactant is the one that is completely consumed and will determine the amount of product formed.
To find the limiting reactant, we need to calculate the ratio of moles of C6H6 to O2 in the balanced equation.
From the balanced equation:
1 mole of C6H6 reacts with 15 moles of O2
Ratio of moles of C6H6 to O2 = 1 / 15
Now, compare the actual ratio of moles of C6H6 to O2:
Actual ratio = Moles of C6H6 / Moles of O2
If the actual ratio is less than the ratio from the balanced equation, C6H6 is the limiting reactant. Otherwise, O2 is the limiting reactant.
Let's calculate the actual ratio:
Actual ratio = (39.3 g / 78.11 g/mol) / (108.3 g / 32.00 g/mol)
Step 3: Calculate the moles of CO2 produced based on the limiting reactant.
If C6H6 is the limiting reactant:
Moles of CO2 = Moles of C6H6 * (6 moles CO2 / 1 mole C6H6)
If O2 is the limiting reactant:
Moles of CO2 = Moles of O2 * (6 moles CO2 / 15 moles O2)
Finally, convert moles of CO2 to grams using the molar mass of CO2.
Molar mass of CO2: (12.01 g/mol) + (2 x 16.00 g/mol) = 44.01 g/mol
Mass of CO2 = Moles of CO2 * Molar mass of CO2
By following these steps, you can calculate the quantity of CO2 produced in the reaction.
To find out how much CO2 will be produced in this reaction, we first need to determine the balanced equation for the reaction between C6H6 (benzene) and O2 (oxygen). The balanced equation is:
C6H6 + 15O2 -> 6CO2 + 3H2O
From the balanced equation, we can see that 1 mole of C6H6 reacts with 15 moles of O2 to produce 6 moles of CO2.
To calculate the moles of C6H6 and O2, we will use the molar masses of the two compounds:
Molar mass of C6H6 = 12.01 g/mol (carbon) + 6 * 1.01 g/mol (hydrogen) = 78.11 g/mol
Molar mass of O2 = 16.00 g/mol (oxygen) * 2 = 32.00 g/mol
Now we can calculate the moles of C6H6 and O2:
Moles of C6H6 = 39.3 g / 78.11 g/mol = 0.503 mol (rounded to three decimal places)
Moles of O2 = 108.3 g / 32.00 g/mol = 3.384 mol (rounded to three decimal places)
Since the balanced equation shows that 1 mol of C6H6 produces 6 mol of CO2, the number of moles of CO2 produced can be calculated as:
Moles of CO2 = (0.503 mol C6H6) * (6 mol CO2 / 1 mol C6H6) = 3.018 mol
Finally, we can convert moles of CO2 to grams by multiplying by the molar mass of CO2:
Mass of CO2 = 3.018 mol * 44.01 g/mol = 132.76 g (rounded to two decimal places)
Therefore, 132.76 grams of CO2 will be produced in this reaction.
The is a limiting reagent (LR) problem and you know that because amounts are given for BOTH reactants.
1. Write and balance the reaction. The products are CO2 and H2O.
2. Convert grams C6H6 (use caps) to mols. mols = grams/molar mass
3. Do the same and convert g O2 to mols.
4. Using the coefficients in the balanced equation, convert mols C6H6 to mols CO2.
5. Do the same and convert mols O2 to mols CO2.
6. It is likely that the values for 2 and 3 will not agree which means one of them is wrong. The correct value in LR problems is ALWAYS the smaller value and the reagent responsible for that value is the LR.
7. Using the smaller value, convert to grams CO2. g = mols x molar mass.