2 C2H6 + 7 O2 → 4 CO2 + 6 H2O
201.6 g C2H6 reacts with 400.0g O2 to make water..
What is the mass of H2O based on the above reactant amounts?
To find the mass of H2O produced, we need to use stoichiometry.
1. Calculate the moles of each reactant:
Moles of C2H6 = mass of C2H6 / molar mass of C2H6
Moles of C2H6 = 201.6 g / (2 * atomic mass of C + 6 * atomic mass of H)
Moles of C2H6 = 201.6 g / (2 * 12.01 g/mol + 6 * 1.008 g/mol)
Moles of C2H6 ≈ 8.34 mol
Moles of O2 = mass of O2 / molar mass of O2
Moles of O2 = 400.0 g / (2 * atomic mass of O)
Moles of O2 = 400.0 g / (2 * 16.00 g/mol)
Moles of O2 ≈ 12.50 mol
2. Determine the limiting reactant:
To determine the limiting reactant, compare the moles of each reactant. The reactant that produces the least amount of product is the limiting reactant.
According to the balanced equation:
2 C2H6 + 7 O2 → 4 CO2 + 6 H2O
The molar ratio of C2H6 to H2O is 2:6 or 1:3.
The molar ratio of O2 to H2O is 7:6 or 7:3.5.
Moles of H2O produced by C2H6 = moles of C2H6 × (6 moles of H2O / 2 moles of C2H6)
Moles of H2O produced by C2H6 = 8.34 mol × (6 / 2)
Moles of H2O produced by C2H6 ≈ 25.02 mol
Moles of H2O produced by O2 = moles of O2 × (6 moles of H2O / 7 moles of O2)
Moles of H2O produced by O2 = 12.50 mol × (6 / 7)
Moles of H2O produced by O2 ≈ 10.71 mol
Since we have fewer moles of H2O produced by O2 (10.71 mol) than C2H6 (25.02 mol), O2 is the limiting reactant.
3. Calculate the mass of H2O produced:
Mass of H2O = moles of H2O produced by O2 × molar mass of H2O
Mass of H2O = 10.71 mol × (2 * atomic mass of H + atomic mass of O)
Mass of H2O = 10.71 mol × (2 * 1.008 g/mol + 16.00 g/mol)
Mass of H2O ≈ 203.51 g
Therefore, the mass of H2O produced based on the given reactant amounts is approximately 203.51 grams.
To find the mass of H2O produced based on the reactant amounts, we need to use stoichiometry. Stoichiometry is a method of calculating the relative quantities of substances involved in a chemical reaction.
First, let's find the mole ratio between C2H6 and H2O from the balanced equation:
2 C2H6 + 7 O2 → 4 CO2 + 6 H2O
According to the equation, for every 2 moles of C2H6, 6 moles of H2O are produced. Therefore, the mole ratio between C2H6 and H2O is 2:6 or 1:3.
Now, let's calculate the number of moles of C2H6 and O2 using their respective molar masses:
Molar mass of C2H6: 2(12.01 g/mol) + 6(1.01 g/mol) = 30.07 g/mol
Number of moles of C2H6 = mass of C2H6 / molar mass of C2H6
= 201.6 g / 30.07 g/mol
≈ 6.71 mol
Molar mass of O2: 2(16.00 g/mol) = 32.00 g/mol
Number of moles of O2 = mass of O2 / molar mass of O2
= 400.0 g / 32.00 g/mol
≈ 12.50 mol
Next, we compare the mole amounts of C2H6 and O2. Based on the balanced equation, the stoichiometric ratio between C2H6 and O2 is 2:7 (2 moles of C2H6 reacts with 7 moles of O2).
Since we have determined that we have 6.71 moles of C2H6 and 12.50 moles of O2, we can now identify the limiting reactant - the reactant that is completely consumed.
To find the limiting reactant, we compare the amounts of moles of C2H6 and O2 using their respective stoichiometric coefficients:
C2H6:O2 ratio = 2:7
(6.71 mol C2H6) / (2 mol C2H6) ≈ 3.35
(12.50 mol O2) / (7 mol O2) ≈ 1.79
From the calculations above, we can see that the mole ratio for C2H6 is greater than for O2. This means that O2 is the limiting reactant because O2 will be completely consumed while there will be some excess C2H6 left after the reaction.
Now that we have identified the limiting reactant, we can calculate the moles of H2O produced using the stoichiometry of the balanced equation:
For every 7 moles of O2, 6 moles of H2O are produced.
Number of moles of H2O = (12.50 mol O2) / (7 mol O2) * (6 mol H2O)
≈ 10.71 mol H2O
Finally, we can calculate the mass of H2O produced using the number of moles and its molar mass:
Molar mass of H2O: 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
Mass of H2O = number of moles of H2O * molar mass of H2O
= 10.71 mol * 18.02 g/mol
≈ 193.18 g
Therefore, based on the given reactant amounts, the mass of H2O produced is approximately 193.18 grams.
This is a limiting reagent (LR) problem. You know that when the amount is given for more than one reactant in the equation.
2 C2H6 + 7 O2 → 4 CO2 + 6 H2O
mols C2H6 = grams/molar mass = 201.6/30 = approx 6.7 but you need to recalculate this AND ALL OF THE OTHERS SINCE I'VE JUST DONE ESTIMATES.
How much H2O would this form? That's mols C2H6 x (6 mols H2O/2 mol C2H6) = about 6.7 x 3 = 20
mols O2 = 400/32 = 12. How much water would that form? That's
12 mols O2 x (6 mols H2O/7 mols O2) = 11
In LR problems the small number wins ; therefore, O2 is the LR and approx 11 mols H2O will be formed. Convert that to grams.
grams H2O = mols H2O x molar mass H2O = ?