Gas stocihiometry... 225.6 L of O2 is allowed to react with 22.4 L C4H10. What is the final volume of gases?

Question

Please help me with that question, giving the way and explain :'(

Thank you,
Ng Yut Sei

Answer 1

2 C4H10 + 13 O2 ==> 8 CO2 + 10 H2O

If you have 22.4 L C4H10, then you would need 6.5 times the amount of O2, which is 145.6 L O2

Since you have way more O2 than you need, the O2 is in excess and the C4H10 is the limiting reactant.

Final volume of gaseous products:

22.4 L x 4 = 89.6 L CO2

22.4 L x 5 = 112 L H2O

You also would have 80 L or unreacted O2 left over.

Answer 2

The first thing you have to do is write a balanced equation:

2 C4H10 + 13 O2 => 8 CO2 + 10 H2O

Since you are not given the conditions, I assume you are at STP. If so, use the conversion factor 1 mole of gas at STP = 22.4 Liters.

Using the conversion factor and the mole ratios, you should be able to take it from there.

Answer 3

Assuming that your measurements are and will be at STP, you have 10 moles of oxygen and 1 mole of butane. The combustion reaction is, on the basis of 1 mole butane in:
.......C4H10 + 6.5 O2 -> 4 CO2 + 5 H2O
So we have enough oxygen to burn the butane, so there isn't any in the final mix. But we will have 3.5 moles oxygen, 4 moles CO2 and 5 moles H2O as vapor. Multiply each by 22.4 L/mole to get the volumes of each and add the volumes to get your desired answer.

Answer 4

first, u need to write out and balance the Chemical equation for this combustion reaction :C4H10 + (9/2)O2---> 2CO2 + 5H2O

mole ratio of gases= volume ratio of gases
So vol ratio of C4H10:O2 =1:4.5
vol of C4H10= 22.4L--->vol of O2 needed = 22.4 x 4.5 =100.8L
since the amount of O2 used(225.6L) > amount of O2 needed for this combustion, C4H10 is the limiting reactant and O2 is the excess reactant.

volume of CO2 produced = 22.4L x mol ratio of CO2:C4H10
= 22.4 x 2/1
= 44.8L
volume of CO2 produced = 22.4L x mol ratio of H2O:C4H10
= 22.4 x 5/1
= 112L
volume of unreacted O2 = 225.6L - 100.8L
= 124.8L
Thus the final volume of gases = 124.8 +112+44.8
=281.6L