C5H12 + O2--- CO2 + H2O give that the equation if I use 85 grams of O2 and excess C5H12 how many grams of CO2 are produced
C5H12 + 8O2--- 5CO2 + 6H2O
mols O2 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols O2 to mols CO2.
Then g CO2 = mols CO2 x molar mass CO2
Balanced equation
C5H12 + 8O2 --> 5CO2 + 6H2O
Well, well, well! Looks like we've got a chemistry problem on our hands! Let's dive right in, shall we?
First things first, we need to figure out the molar ratio between the reactants and the product. According to the balanced equation:
1 mole of C5H12 reacts with 8 moles of O2 to produce 5 moles of CO2.
Since we're given 85 grams of O2, let's convert that to moles. The molar mass of O2 is approximately 32 grams/mol:
85 grams of O2 * (1 mol O2/32 grams O2) ≈ 2.65625 moles of O2
Based on the molar ratio, we see that 8 moles of O2 are needed to react with 1 mole of C5H12. However, since we have an excess of C5H12, we don't need to worry about it limiting the reaction.
Now, let's calculate the amount of CO2 produced using the mole ratio. 1 mole of C5H12 produces 5 moles of CO2:
2.65625 moles of O2 * (5 moles CO2/8 moles O2) ≈ 1.66015625 moles of CO2
Now, convert moles of CO2 to grams. The molar mass of CO2 is approximately 44 grams/mol:
1.66015625 moles CO2 * (44 grams CO2/1 mol CO2) ≈ 72.890625 grams of CO2
So, approximately 72.89 grams of CO2 will be produced. I hope this answer brought you chemistry cheer!
To find the amount of CO2 produced, we need to use the balanced equation and the given amount of oxygen (O2) in grams.
Let's start by balancing the equation:
C5H12 + O2 → CO2 + H2O
To balance the equation, we need to have the same number of each element on both sides. Counting the carbons, hydrogens, and oxygens, we see that there are 5 carbon atoms, 12 hydrogen atoms, and 2 oxygen atoms in the reactant (C5H12). On the product side, there's 1 carbon atom, 2 oxygen atoms, and 2 hydrogen atoms. We can balance the equation by multiplying the reactant side as follows:
C5H12 + 8O2 → 5CO2 + 6H2O
Now that we have a balanced equation, we can use stoichiometry to find the amount of CO2 produced.
The molar ratio between C5H12 and CO2 is 1:5, according to the balanced equation. This means that for every 1 mole of C5H12, 5 moles of CO2 are produced.
To find the moles of O2, we use the molar mass of oxygen, which is approximately 32 g/mol. Given that we have 85 grams of O2, we can calculate the number of moles:
Moles of O2 = Mass of O2 / Molar mass of O2
Moles of O2 = 85 g / 32 g/mol
Moles of O2 ≈ 2.66 mol
Since the reaction is balanced at a 1:5 ratio of C5H12 to CO2, the number of moles of CO2 produced will be 5 times the moles of C5H12. However, the excess C5H12 is not specified, so we cannot calculate the exact amount of CO2 produced.
Nevertheless, we can still determine the number of moles of CO2 produced using the molar ratio. Without the actual amount of excess C5H12, we cannot obtain the exact grams of CO2 produced.