Question

if 3 moles of hydrogen burn with excess oxygen, how many liters of water vapor will form?

Answer 1

First, write a balanced equation for the reaction. Next, can you find the number of moles of whatever substance you have?

Use this: http://www.ptable.com/

Answer 2

_3H + _O 2 -> _ H2O

Answer 3

not sure how to find the moles...

Answer 4

That is not balanced. Can you balance it first?

Answer 5

4 3H + 2 O2 ---> 6 H2O

Answer 6

Your oxygen is not balanced. And hydrogen does not exist as single molecules. They are always diatomic. It is H2 not H

Answer 7

Opps! I totally forgot! its 2 3H2 + 3 O2 ----> 6 H2O

Answer 8

Yup looks right to me. Now, the number of moles is the mass of a substance divide by its molar mass. Can you do that?

Answer 9

Which reactant or product am i finding the moles for?

Answer 10

You are only given the mass of hydrogen which is the reactant.

Answer 11

i'm still confused... We given the mass of Oxygen as will in the balanced equation? If we were originally given 3 moles would that just be it...?

Answer 12

Ooh you weren't given the mass, you were given the number of moles of hydrogen. So you're looking for the volume of H2O. So from the balanced equation, you can see that portions of H2 will produce 6 portions of H2O. How would you get the number of moles of H2O then?

Answer 13

would it be the overall mass converted to moles? 203.9748 * ( 1 mol / 6.002 *10^23 )

Answer 14

Duuuude. We didn't even balance the equation correctly. Ok new equation what @chmvijay said.

Answer 15

2H2+O2 ------>2H2O
when two moles of H2 react with excess oxygen it produces 2 moles of water (by above formula)
so for 2 moles of H2 produces 2 moles of water
so 3 moles of water produces = 3 moles of water
1 mole of water= 16 g of water
3 mole of water = 16 *3 =48 g of water
density = mass /volume u do and get the volume

Answer 16

Wait we aren't given the density. I think we're supposed to use STP

Answer 17

density of water is almost =1
its basic no need to mention

Answer 18

@lelenall did u get what i said above ?

Answer 19

Yesss!!!! That was actually so helpful!!!! Thank you!!!! :))

Answer 20

@chmvijay

Answer 21

ur welcome :)