Question

Question #ede60

Answer 1

22 grams of carbon dioxide.

The thing you need to know here is called the Law of Mass Conservation. Basically it says that, because all the same atoms are there before and after the reaction (just bonded together in different ways), then the total mass of the products must be the same of the total mass of the reactants.

In the example you gave, you had 8g of methane and 32g of oxygen as your reactants: a total of 40g. This means the products together weight 40g. The water produced weighs 18g, so the carbon dioxide produced should weigh 40-18=22g.

Just to check, we ought to make sure that 8g of methane actually does produce 18g of water, and react with 32g of oxygen.
The reaction is `CH_4 + 2O_2 -> CO_2 + 2H_2O`

Moles of methane = mass of methane / `M_r` of methane
= 8 / 16 = 0.5
Moles of oxygen this will react with = 2 x moles of methane (from balanced equation) = 2 x 0.5 = 1

Mass of oxygen = moles of oxygen x `M_r` of `O_2` = 1 x 32 = 32g
So that part of the question checks out.

In the same way, moles of water = 2 x moles of methane (from balanced equation) = 2 x 0.5 = 1

Mass of water = moles of water x `M_r` of `H_2O` = 1 x 18 = 18g
So that part checks out too.

Using the same method rather than mass conservation, we can work out that moles of carbon dioxide = moles of methane (from balanced equation) = 0.5

Mass of carbon dioxide = moles of `CO_2` x `M_r` of `CO_2` = 0.5 x 44 = 22g which matches the answer we got by the other method, and confirms that neither reactant was in excess.