Pentane (C5H12) burns in oxygen to produce carbon dioxide and water.
C5H12(g)+8O2(g)→5CO2(g)+6H2O(g)
Calculate the mass of CO2 that can be produced if the reaction of 38.5 g of pentane and sufficient oxygen has a 58.0 % yield.
What was the answer exactly?
CO and H2O
mols pentane = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols pentane to mols CO2.
Then g CO2 = mols CO2 x molar mass CO2 and this is the mass at 100% yield.
Mass CO2 at 58.0% yield is mass at 100% yield x 0.580 = ?
Sodium bicarbonate in an reacts with hydrochloric acid to form carbon dioxide sodium chloride and water
Alright, let's calculate the mass of CO2 produced!
First, we need to determine the molar mass of pentane (C5H12). To do this, we multiply the molar masses of carbon (12.01 g/mol) and hydrogen (1.01 g/mol) by their respective subscripts and sum them up:
(5 * 12.01 g/mol) + (12 * 1.01 g/mol) = 72.15 g/mol
Now, we can calculate the number of moles of pentane by dividing the given mass (38.5 g) by the molar mass:
38.5 g / 72.15 g/mol = 0.534 mol
According to the balanced equation, 1 mole of pentane yields 5 moles of CO2. Therefore, we can calculate the theoretical yield of CO2 by multiplying the number of moles of pentane by the stoichiometric ratio:
0.534 mol * 5 mol CO2/1 mol C5H12 = 2.67 mol CO2
Since the yield is given to be 58.0%, we multiply the theoretical yield by the percentage yield:
2.67 mol CO2 * 0.58 = 1.55 mol CO2
Finally, we can calculate the mass of CO2 produced by multiplying the number of moles by the molar mass of CO2:
1.55 mol * (44.01 g/mol) = 68.11 g
Therefore, the mass of CO2 that can be produced with a 58.0% yield from the reaction of 38.5 g of pentane and sufficient oxygen is approximately 68.11 g. Now, go forth and embrace the gassy humor of carbon dioxide!
To calculate the mass of CO2 produced, we need to follow a few steps:
Step 1: Calculate the molar mass of pentane (C5H12).
The molar mass of pentane can be calculated by summing the atomic masses of its constituent elements, carbon (C) and hydrogen (H).
Molar mass of C5H12 = (molar mass of C) * 5 + (molar mass of H) * 12.
Looking up the atomic masses of carbon and hydrogen, we find:
Molar mass of C = 12.01 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of C5H12 = (12.01 g/mol) * 5 + (1.01 g/mol) * 12.
Step 2: Calculate the moles of pentane (C5H12).
We can use the given mass of pentane (38.5 g) and its molar mass to determine the number of moles.
Moles of C5H12 = Mass of C5H12 / Molar mass of C5H12.
Step 3: Determine the limiting reagent.
We need to identify the limiting reagent, which is the reactant that will be completely consumed in the reaction. In this case, we compare the moles of pentane (C5H12) with the stoichiometry of the reaction.
From the balanced equation 1 mol of C5H12 reacts with 5 mol of CO2.
So, moles of CO2 produced = Moles of C5H12 * (5 mol CO2 / 1 mol C5H12).
Step 4: Calculate the mass of CO2 produced.
To calculate the mass of CO2 produced, we multiply the moles of CO2 by its molar mass.
Mass of CO2 = Moles of CO2 * Molar mass of CO2.
Now, let's calculate the values step by step:
Step 1: Calculate the molar mass of pentane (C5H12):
Molar mass of C5H12 = (12.01 g/mol * 5) + (1.01 g/mol * 12).
Step 2: Calculate the moles of pentane (C5H12):
Moles of C5H12 = 38.5 g / Molar mass of C5H12.
Step 3: Determine the limiting reagent:
Moles of CO2 produced = Moles of C5H12 * (5 mol CO2 / 1 mol C5H12).
Step 4: Calculate the mass of CO2 produced:
Mass of CO2 = Moles of CO2 produced * Molar mass of CO2.
Note: The given yield of 58.0% will be applied after calculating the theoretical mass of CO2 produced.
I hope this helps! Let me know if you need any further assistance with the calculations.