When 0.340 mol of butane, C4H10, are burned with excess oxygen giving CO2 and H2O, how many moles of oxygen are consumed?
balance the equation. I bet you get four moles of CO2 and 5 moles of water for each mole of butane. If that is true, then you must use 4O2+5(1/2O2) of oxygen, or 6.5 moles O2 for each mole of butane. So you had .340moles, and that multiplied by 6.5 is ...
Balance the equation and see
wrong
12
13.0
Well, let's do some combustion comedy!
Why did the butane go to the party with excess oxygen?
Because it wanted to ignite some hot chemistry with CO2 and H2O!
Now, let's talk about moles.
In every mole of butane, we have 10 moles of hydrogen. To burn 0.340 mol of butane, we will need 3.40 moles of hydrogen.
And since we have 2 moles of hydrogen in every mole of oxygen, we'll need twice that amount, which is 6.80 moles of oxygen.
So, the answer is 6.80 moles of oxygen are consumed in this fiery reaction!
Hope this answer sparked some laughter while you learned about moles and gas reactions!
To determine the number of moles of oxygen consumed in the combustion of butane, we first need to balance the chemical equation.
The balanced combustion equation for butane (C4H10) with excess oxygen (O2) is:
2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O
From the balanced equation, we can see that 2 moles of butane react with 13 moles of oxygen to produce 8 moles of carbon dioxide and 10 moles of water.
Given that we have 0.340 mol of butane, we need to determine the number of moles of oxygen consumed.
Using the ratio from the balanced equation, we can set up a proportion:
2 mol of C4H10 / 13 mol of O2 = 0.340 mol of C4H10 / x mol of O2
Cross-multiplying, we get:
2 * x = 13 * 0.340
Simplifying:
2x = 4.42
Dividing both sides by 2:
x = 2.21
Therefore, approximately 2.21 moles of oxygen are consumed in the combustion of 0.340 mol of butane.