3. When a mixture of 10.0 grams of acetylene (C2H2) and 10.0 grams of oxygen (O2) is ignited, the resulting combustion reaction produces CO2 and H2O.

A) Write a balanced equation for this reaction.
B) What is the limiting reactant?
C) How many grams of C2, H2, O2, CO2, and H2O are present after the reaction is complete?

2C2H2 + 5O2 ==> 4CO2 + 2H2O

b.
mols C2H2 = grams/molar mass
mols O2 = grams/molar mass
Using the coefficients in the balanced equation, convert mols C2H2 to mols CO2.
Do the same for mols O2.
It is likely that the two values for mols CO2 will not be the same which means one of them is wrong. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that number is the limiting reagent. Then convert grams to mols by g = mols x molar mass.

c.
After you have determined the limiting reagent, use that value and the procedure above to calculate mols CO2 and mols H2O formed.
Use the same procedure using the limiting reagent to determine the mols of the non-limiting reagent used and subtract from the initial amount to determine the value after the reaction.

A) The balanced equation for this reaction is:

2C2H2 + 5O2 -> 4CO2 + 2H2O

B) To find the limiting reactant, we need to compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation.

Molar mass of C2H2 = 2(12.01 g/mol) + 2(1.01 g/mol) = 26.04 g/mol
Molar mass of O2 = 2(16.00 g/mol) = 32.00 g/mol

Number of moles of C2H2 = 10.0 g / 26.04 g/mol = 0.384 mol
Number of moles of O2 = 10.0 g / 32.00 g/mol = 0.313 mol

According to the balanced equation, the stoichiometric ratio between C2H2 and O2 is 2:5. Therefore, for every 1 mole of C2H2, we need 2.5 moles of O2.

Ratio of moles of C2H2 to O2 = 0.384 mol / 0.313 mol = 1.23

Since this ratio is less than the stoichiometric ratio of 2.5, it means that C2H2 is the limiting reactant.

C) To find the amounts of each compound after the reaction, we need to use stoichiometry.

Number of moles of C2H2 used = 0.384 mol (limiting reactant)
Number of moles of O2 used = (2/5) x 0.384 mol = 0.1536 mol

Number of moles of CO2 produced = 2 x 0.384 mol = 0.768 mol
Number of moles of H2O produced = 1 x 0.384 mol = 0.384 mol

To convert the moles of each compound to grams, we can use their respective molar masses.

Molar mass of CO2 = 12.01 g/mol + 2(16.00 g/mol) = 44.01 g/mol
Molar mass of H2O = 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol

Mass of CO2 produced = 0.768 mol x 44.01 g/mol = 33.78 g
Mass of H2O produced = 0.384 mol x 18.02 g/mol = 6.92 g

After the reaction is complete:
- The mass of C2H2 remaining will be 10.0 g - mass of C2H2 used.
- The mass of O2 remaining will be 10.0 g - mass of O2 used.
- The mass of CO2 and H2O will be the masses produced during the reaction.

Please note that the calculation for the remaining masses will depend on the original amounts of C2H2 and O2 present in the mixture.

A) To write a balanced equation for the combustion reaction between acetylene (C2H2) and oxygen (O2), we need to ensure that the number of atoms in the reactants is equal to the number of atoms in the products.

The balanced equation for the reaction is:

2 C2H2 + 5 O2 -> 4 CO2 + 2 H2O

B) The limiting reactant is determined by comparing the stoichiometry of the reactants. The reactant that is completely consumed and limits the amount of product formed is the limiting reactant.

To find the limiting reactant, we need to calculate the number of moles for each reactant by dividing the given mass by their respective molar masses.

Given:
Mass of acetylene (C2H2) = 10.0 grams
Mass of oxygen (O2) = 10.0 grams

The molar mass of acetylene (C2H2) = (2 * atomic mass of carbon) + (2 * atomic mass of hydrogen)
= (2 * 12.01 g/mol) + (2 * 1.01 g/mol)
= 26.04 g/mol

The molar mass of oxygen (O2) = 2 * atomic mass of oxygen
= 2 * 16.00 g/mol
= 32.00 g/mol

Number of moles of acetylene (C2H2) = mass / molar mass
= 10.0 g / 26.04 g/mol
≈ 0.384 moles

Number of moles of oxygen (O2) = mass / molar mass
= 10.0 g / 32.00 g/mol
≈ 0.312 moles

Comparing the coefficients of the balanced equation, we see that the ratio of acetylene to oxygen is 2:5. Therefore, every 2 moles of C2H2 requires 5 moles of O2.

The ratio of moles of acetylene to oxygen is 0.384/0.312 = 1.23.
Since the ratio is less than 2/5, we can conclude that oxygen (O2) is the limiting reactant.

C) To determine the number of moles and grams of each substance present after the reaction is complete, we will use the stoichiometry of the balanced equation.

According to the balanced equation:
2 moles of C2H2 produce 4 moles of CO2
2 moles of C2H2 produce 2 moles of H2O
5 moles of O2 produce 10 moles of CO2
5 moles of O2 produce 2 moles of H2O

To calculate the number of moles of CO2 and H2O produced:
Number of moles of CO2 = 4 moles (from C2H2) + 10 moles (from O2)
= 4 moles + 10 * (0.312 moles of O2, which was the limiting reactant)
= 4 moles + 3.12 moles
= 7.12 moles

Number of moles of H2O = 2 moles (from C2H2) + 2 moles (from O2)
= 2 moles + 2 * (0.312 moles of O2, which was the limiting reactant)
= 2 moles + 0.624 moles
= 2.624 moles

To calculate the number of moles of C2H2 remaining:
Number of moles of C2H2 remaining = 0.384 moles (initial amount) - 2 moles (consumed)
= -1.616 moles

Since the number of moles cannot be negative, we conclude that there is no C2H2 remaining after the reaction.

To calculate the number of moles of O2 remaining:
Number of moles of O2 remaining = 0.312 moles (initial amount) - 0.312 moles (consumed)
= 0 moles

Therefore, there is no O2 remaining after the reaction.

To calculate the grams of CO2 and H2O produced:
Mass of CO2 = number of moles of CO2 * molar mass of CO2
= 7.12 moles * (atomic mass of carbon + 2 * atomic mass of oxygen)
= 7.12 moles * (12.01 g/mol + 2 * 16.00 g/mol)
≈ 432.92 grams

Mass of H2O = number of moles of H2O * molar mass of H2O
= 2.624 moles * (2 * atomic mass of hydrogen + atomic mass of oxygen)
= 2.624 moles * (2 * 1.01 g/mol + 16.00 g/mol)
≈ 92.08 grams

Therefore, after the reaction is complete, there are 432.92 grams of CO2 and 92.08 grams of H2O present.