Using the equation:
2C2H6 + 7O2--> 4CO2 + 6H20
How many moles of CO2 are produced by the reaction of 21.0 mol O2?
Again, not sure how to write it correctly, but if I multiply 7O2 by 3, then I should do the same to the other side of the equation to make it 12CO2?
that is right.
But why don't you learn to do it right so when you have a number that isn't an even multiple, you will know how to do it. For this problem, that is
21 mols O2 x (4 mols CO2/7 mols O2) = 21 x 4/7 = 12
To determine the number of moles of CO2 produced by the reaction of 21.0 mol O2, we'll start by setting up a mole ratio using the stoichiometric coefficients in the balanced equation:
2C2H6 + 7O2 → 4CO2 + 6H2O
The mole ratio between O2 and CO2 is 7:4, which means for every 7 moles of O2 consumed, 4 moles of CO2 are produced.
To find the mole amount of CO2 produced, we can use the following conversion:
Moles of O2 (given) * (moles of CO2 / moles of O2) = Moles of CO2
Using the ratio from the equation (7 moles O2: 4 moles CO2), we can calculate:
21.0 mol O2 * (4 mol CO2 / 7 mol O2) = 12.0 mol CO2
Therefore, the reaction of 21.0 mol O2 produces 12.0 mol CO2. The correct value for the moles of CO2 produced is not 12CO2, but rather 12.0 mol CO2.
To determine how many moles of CO2 are produced by the reaction of 21.0 mol O2, you need to use the balanced equation you provided:
2C2H6 + 7O2 -> 4CO2 + 6H2O
The coefficients in the balanced equation represent the mole ratios between the reactants and products. In this case, the coefficient of CO2 is 4, which means that for every 7 moles of O2 reactant, 4 moles of CO2 are produced.
To find the number of moles of CO2 produced by the reaction of 21.0 mol O2, you can set up a proportion using the mole ratios:
(4 moles CO2 / 7 moles O2) = (x moles CO2 / 21.0 moles O2)
Cross-multiplying and solving for x gives:
x = (4 moles CO2 / 7 moles O2) * 21.0 moles O2
x = 12 moles CO2
Therefore, 21.0 mol O2 will produce 12.0 mol CO2, not 12CO2.
Remember, when balancing an equation, you can change the coefficients but not the subscripts within the compounds.