Chem review help?

Question

I am doing a review for AP Chem, which I am about to begin, and I am stumped on the method of which to do this, it has been too long since i have done any chem.

3a. Write a combustion reaction of methane (CH4).

b. If 10.0 g of methane reacts with 25.0 g of oxygen, how many grams water will be formed?

c. How many grams of the excess reactant will remain having not reacted?

d. If 9.00 g of water is produced, what is the % yield of the reaction?

Answer 1

a) CH4   +  2O2   -->   CO2    +  2H2O

always add oxygen when its combustion and products shud be water & carbon dioxide

b) Convert to moles using moles = Mass/Molar Mass

and use stochiometry

c) Using the reactant with less number of moles, again using stochiometry calculate how much of the other reactant is required and then subtract from the given amount

d) % yield = Actual yield/theoretical yield * 100

Answer 2

CH4 + 2O2 ---->  CO2  + 2H2O

64g of O2 reacts with 16g of CH4

25g      ===========   25*16/64 = 6.25gof CH4

If 16g CH4 produces 36g of H2O

6.25 ============= 6.25*36/16 = 14.0625g H2O

unreacted 10-6.25 = 3.75g of CH4

If 9g of water is produced % yield = 9*100/14.0625 =64%

Answer 3

a.

CH4 + 2O2  ---->   2H2O   + CO2

All combustion reactions follow this formula when presented under excess oxygen conditions. Combustion reactions are very exothermic radical reactions but have a high activation energy.

Sometimes a combustion reaction can occur under limited oxygen in which case this would be the reaction.

2CH4 + 3O2 (limited)  ---->   4H2O   + 2CO

The only difference is that Carbon Monoxide (CO) is evolved instead of CO2 (Carbon Dioxide). This sort of a reaction takes place in cars (except  other hydrocarbons) which is why cars exhaust CO.

b.

Usually just combustion reaction assume excess O2 so we use the first combustion reaction:

CH4 + 2O2  ---->   2H2O   + CO2

We need to find whether the Methane of the Oxygen is the limiting reagent. That is, if the reaction runs out of either Oxygen or Methane first, then the reaction will stop so we need to see which will run out first.

10.0g CH4  * 1n/ (12 + 4) = 0.625 n CH4

25.0g O2 * 1n/ (16 + 16) = 0.781 n O2

So if we react 0.625  n CH4 we require 1.25 n O2 but we just calculated that we only have 0.781 n O2, so O2 is our limiting reagent so we use it to calculate the grams of water formed.

0.781 n O2  *  2n H2O/ 2n O2 = 0.781 n H20

0.781n H20 * (18.0 g/n) = 14.1 g H20

c. So Methane is our excess, we reacted 0.781 n O2 and would require only  0.781 n 02 * 1n CH4/2n O2  = 0.391 n CH4  and we started with 0.625 n CH4

0.625 n - 0.391n = 0.234 n CH4 left over.

0.234 n CH4 * (16g CH4/n)  = 3.75 g CH4 remains

d.

The ideal amount or theoretical yield of H20 we calculated above at 14.1 g H20

The actual yield is said to be 9.00 g

(Actual Yield/Theoretical Yield)  * 100 = %yeild

%yield = (9.00/14.1)  * 100   = 63.8%

Not too bad of a yield

hope that helped.