The combustion of a hydrocarbon to produce carbon dioxide and water is a 5th type of chemical reaction.
A fun demonstration is making bubbles of methane gas and lighting them -- a taper or candle taped to a meter stick can be used to light the bubbles up high, away from observers. Lab goggles are required!
A fun demonstration is making bubbles of methane gas and lighting them -- a taper or candle taped to a meter stick can be used to light the bubbles up high, away from observers. Lab goggles are required!
Notice that he is using a funnel attached a rubber hose which is attached to the gas outlet. The funnel is dipped in soap solution and when the gas is turned on, a bubble can be formed. BTW, it is more difficult to form the bubbles during dry weather (such as the winter months).
One small challenge of the reaction is balancing -- sometime it's easy (with a hydrocarbon with an even number of carbons):
C2H6 + O2 --> CO2 + H2O
Because the oxygen is alone on the reactants side and split on the products side, and reactant oxygen is probably going to be the last coefficient entered to balance the reaction.
C2H6 +____O2 --> 2 CO2 + ___ H2O
C2H6 + ____ O2 --> 2 CO2 + 3 H2O
4 + 6 = 10
C2H6 + 5 O2 --> 2 CO2 + 3 H2O
If there is an odd number of carbons (e.g. 1, 3, 5, etc.) then you would need a fractional coefficient for the reactant oxygen and then double all of the coefficients (to get rid of the fraction).
A trick I taught my students was to automatically double the hydrocarbon coefficient -- this will lead to a whole number coefficient for the reactant oxygen:
C3H8 + O2 --> CO2 + H2O
2 C3H8 + __ O2 --> __ CO2 + __ H2O
2 C3H8 + __ O2 --> __ CO2 + 8 H2O
2 C3H8 + ___ O2 --> 6 CO2 + 8 H2O
12 + 8 = 20
2 C3H8 + 10 O2 --> 6 CO2 + 8 H2O
But wait! The coefficients are not in the lowest ratio! They all can be divided by two!
C3H8 + 5 O2 --> 3 CO2 + 4 H2O
Please let me know about your experiences, ask if you have any questions or if you have ideas for other topics for this blog.
And, check out my lab book "Chemistry on a Budget" at:
http://www.amazon.com/Chemistry-Budget-Marjorie-R-Heesemann/dp/0578129159/ref=sr_1_1?s=books&ie=UTF8&qid=1389410170&sr=1-1&keywords=chemistry+on+a+budget
Have a good week!
One small challenge of the reaction is balancing -- sometime it's easy (with a hydrocarbon with an even number of carbons):
C2H6 + O2 --> CO2 + H2O
Because the oxygen is alone on the reactants side and split on the products side, and reactant oxygen is probably going to be the last coefficient entered to balance the reaction.
C2H6 +____O2 --> 2 CO2 + ___ H2O
C2H6 + ____ O2 --> 2 CO2 + 3 H2O
4 + 6 = 10
C2H6 + 5 O2 --> 2 CO2 + 3 H2O
If there is an odd number of carbons (e.g. 1, 3, 5, etc.) then you would need a fractional coefficient for the reactant oxygen and then double all of the coefficients (to get rid of the fraction).
A trick I taught my students was to automatically double the hydrocarbon coefficient -- this will lead to a whole number coefficient for the reactant oxygen:
C3H8 + O2 --> CO2 + H2O
2 C3H8 + __ O2 --> __ CO2 + __ H2O
2 C3H8 + __ O2 --> __ CO2 + 8 H2O
2 C3H8 + ___ O2 --> 6 CO2 + 8 H2O
12 + 8 = 20
2 C3H8 + 10 O2 --> 6 CO2 + 8 H2O
But wait! The coefficients are not in the lowest ratio! They all can be divided by two!
C3H8 + 5 O2 --> 3 CO2 + 4 H2O
Please let me know about your experiences, ask if you have any questions or if you have ideas for other topics for this blog.
And, check out my lab book "Chemistry on a Budget" at:
http://www.amazon.com/Chemistry-Budget-Marjorie-R-Heesemann/dp/0578129159/ref=sr_1_1?s=books&ie=UTF8&qid=1389410170&sr=1-1&keywords=chemistry+on+a+budget
Have a good week!
RSS Feed