How many grams of water will be produced in the complete combustion of 20.0g of Octane, C8H18(g) and 50.0g Oxygen gas?
Combustion reactions always follow this general reaction:
CxHy + O2 -----> CO2 + H2O
Plug in Octane into the the reaction and then balance:
C8H18 + O2 -----> CO2 + H2O
2 C8H18 + 25 O2 -----> 16 CO2 + 18 H2O
Just eyeballing it, octane is the limiting reagent.
****To determine the limiting reagent, you have to find the number of moles that each reactant can produce (i.e., which reactant will you run out of first.)
20.0g *(1 mole/114.23 g)= moles of C8H18
moles of C8H18*(18 mol of H2O/2 moles of C8H18)= moles of H2O
moles of H2O*(18.00g/mole)= mass of H2O
28.36
To determine the number of grams of water produced, we first need to balance the equation for the combustion of octane:
C8H18 + 12.5O2 → 8CO2 + 9H2O
From the balanced equation, we can see that for every 1 mole of octane (C8H18), we get 9 moles of water (H2O).
Now, let's calculate the number of moles of octane (C8H18) and oxygen (O2) given the masses provided:
Molar mass of octane (C8H18) = (12.01 x 8) + (1.01 x 18) = 114.23 g/mol
Molar mass of oxygen (O2) = (16.00 x 2) = 32.00 g/mol
Moles of octane (C8H18) = Mass of octane / Molar mass of octane
= 20.0 g / 114.23 g/mol
≈ 0.175 mol
Moles of oxygen (O2) = Mass of oxygen / Molar mass of oxygen
= 50.0 g / 32.00 g/mol
≈ 1.563 mol
Since the balanced equation tells us that 1 mole of octane yields 9 moles of water, we can now calculate the moles of water produced.
Moles of water = Moles of octane x (9 moles of water / 1 mole of octane)
= 0.175 mol x (9/1)
= 1.575 mol
Finally, let's convert the moles of water into grams:
Mass of water = Moles of water x Molar mass of water
= 1.575 mol x (18.015 g/mol)
≈ 28.35 g
Therefore, approximately 28.35 grams of water will be produced in the complete combustion of 20.0 grams of octane and 50.0 grams of oxygen gas.
To determine the amount of water produced in the complete combustion of Octane and Oxygen gas, you need to first balance the chemical equation and then use stoichiometry.
1. Balance the chemical equation:
C8H18 + 12.5 O2 --> 8 CO2 + 9 H2O
2. Calculate the molar masses:
The molar mass of Octane (C8H18) = 8(12.01 g/mol) + 18(1.01 g/mol) = 114.23 g/mol
The molar mass of Oxygen (O2) = 2(16.00 g/mol) = 32.00 g/mol
The molar mass of Water (H2O) = 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
3. Convert the given masses to moles:
Moles of Octane = 20.0 g / 114.23 g/mol = 0.175 mol
Moles of Oxygen = 50.0 g / 32.00 g/mol = 1.5625 mol
4. Determine the limiting reactant:
To find the limiting reactant, compare the moles of Octane to Oxygen using the stoichiometry coefficients from the balanced equation. The ratio of Octane to Oxygen is 1:12.5.
Moles of Octane / Stoichiometry coefficient of Octane = 0.175 mol / 1 = 0.175 mol
Moles of Oxygen / Stoichiometry coefficient of Oxygen = 1.5625 mol / 12.5 = 0.125 mol
Since the ratio of Octane to Oxygen is 1:12.5, we can see that Oxygen is the limiting reactant as it is present in lesser amount.
5. Calculate moles of Water produced:
Using the stoichiometry coefficients from the balanced equation, we can determine the number of moles of Water produced.
Moles of Oxygen * Stoichiometry coefficient of Water = 0.125 mol * 9 = 1.125 mol
6. Calculate the mass of water produced:
Mass of Water = Moles of Water * Molar mass of Water
Mass of Water = 1.125 mol * 18.02 g/mol = 20.26 g
Therefore, in the complete combustion of 20.0g of Octane and 50.0g of Oxygen gas, 20.26 grams of Water will be produced.