1=what is the percntage yeild of carbon dioxide if 4 moles of octane are burnt to produce 1350g of carbon dioxide
2c8h18 + 25o2 = 16co2 + 18h2o
You really need to find the caps key and use it. o2 doesn't man anything to me. Example. CO is carbon monoxide; Co is cobalt; co is the abbreviation for company.
2C8H18 + 25O2 = 16CO2 + 18H2O
How much CO2 should be produced by 4 moles octane at 100%? That's 4 mols C8H18 x (16 mols CO2/2 mols C8H18) = 4*16/2 = 32 mols CO2.
How many grams is that? That's 32 mols CO2 x 44 g/mol = ? grams CO2. That is the theoretical yield (TY).
The actual yield (AY) is given in the problem as 1350 g.
% yield = (AY/TY)100 = ?
To find the percentage yield of carbon dioxide, we need to determine the actual yield and the theoretical yield.
The actual yield is given in the problem as 1350g of carbon dioxide.
To calculate the theoretical yield of carbon dioxide, we need to use the balanced equation and the molar mass of octane and carbon dioxide.
Step 1: Calculate the molar mass of octane (C8H18):
C = 12.01 g/mol
H = 1.01 g/mol
C8H18 = (12.01 g/mol x 8) + (1.01 g/mol x 18) = 114.23 g/mol
Step 2: Convert the given 1350g of carbon dioxide to moles:
Molar mass of CO2 = (12.01 g/mol x 1) + (16.00 g/mol x 2) = 44.01 g/mol
Moles of CO2 = 1350g CO2 / 44.01 g/mol = 30.66 mol CO2
Step 3: Determine the stoichiometric ratio between octane and carbon dioxide from the balanced equation:
2 C8H18 + 25 O2 → 16 CO2 + 18 H2O
From the equation, we see that for every 2 moles of octane burned, 16 moles of carbon dioxide are produced. This means that the theoretical yield is directly proportional to the moles of octane.
Step 4: Calculate the theoretical yield of carbon dioxide:
Moles of CO2 (theoretical) = (30.66 mol CO2 / 2 mol octane) x 16 mol CO2 = 245.28 mol CO2
Step 5: Calculate the percentage yield:
Percentage yield = (Actual yield / Theoretical yield) x 100
Percentage yield = (30.66 mol CO2 / 245.28 mol CO2) x 100 = 12.5%
Therefore, the percentage yield of carbon dioxide is 12.5%.