Question

16 g of CH4 react with 64 g of O2 producing 44 g of CO2 how many grams of water are produced?

Answer 1

Start by doing a ballanced chemical equation

Answer 2

Sorry, I don´t know what that is

Answer 3

CH4 + O2 --> CO2 + H20 would be the unballanced chemical equation, Can you balance it?

Answer 4

does that mean adding?

Answer 5

that means making sure that the amount of atoms on the right and the left add up

Answer 6

but how do i know if they add up?

Answer 7

CH4 + 2 O2 --> CO2 + 2 H2O you have to count up the atoms

Answer 8

now you have to figure out the limiting reagent and use some stoicemetry to find the mass of water

Answer 9

ok

Answer 10

sorry, but i really don´t know this stuff

Answer 11

Have you ever heard of any of the terms i'm using?

Answer 12

Only of balancing

Answer 13

\[16 g CH_4 (\frac{1 mole CH_4}{16.04 g CH_4})(\frac{2 moles H_2O}{1 mole CH_4})(\frac{18.02 g H_2O}{1 mole H_20})=35.95 g H_2O\]

then we have to do another factor lable for O2 to see if it produces more or less than the CH4, also you might want to check my molar masses due to the fact i'm doing them in my head.

Answer 14

\[64 g O_2 (\frac{1 mole O_2}{32 g O_2})(\frac{2 mole H_20}{2 mole O_2})(\frac{18.02 g H_2O}{1 mole H_2O})=36.04 g H_2O\]

Answer 15

Now which one do you think is going to be the limiting reagent?

Answer 16

the 2nd one

Answer 17

Well it is the first one but whats your reasoning for thinking its the 2nd one?

Answer 18

I thought the process looked a bit more accurate

Answer 19

Well it is possible I messed up on the molar mass doing them from my head but when we are attempting to find the limiting reagent we look for the one that has the smaller number after we complete the factor lable. This is the same as asking which of your reagents is going to run out first.

Answer 20

Does that help?

Answer 21

yes, tnx!

Answer 22

glad I could help keep up the good work