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OpenSessame
Ammonia, NH3, is produced through a combination reaction involving N2(g) and H2(g). If 24.0 mol of H2(g) react with excess N2(g), how many moles of ammonia are produced?
I need to know how to solve this, I'm a bit rusty on my moles.
First, make sure the equation is balanced. Next, can you find the number of moles of whatever substance you have? Use this: http://www.ptable.com/
I mean, write a balanced equation first.
So..\[N _{2}(g) + h _{2}(g) \rightarrow NH _{3}\]
Is that the right equation?
\[N _{2}+3H _{2} \rightarrow 2NH _{3}\]
Oh my the equation editor is not working for me but I can see it. From here can you just use normal typing? Notice that the equation is not balanced. You have 2 N on the left and one on the right. And you have 2 H on the left and 3 on the right.
Right. So now, you know how much H2 you used. Can you fine the number of moles of it? http://erhs.rcsnc.org/UserFiles/Servers/Server_4766394/Image/Donna%20Scofield/mole%20conversion%20map.bmp
Uh...Walk me through?
Sure. Do you know how to find the number of moles?
I know a mole is 6.02 * 10^23 thats about it
That is the Avogadro number - the number of objects in 1 mole of substance. But I'm sorry I misread the question. You are already given the number of moles. So here, from our balanced equation N2 + 3H2 = 2NH3 we can see that 3 parts of H2 produces 2 parts of NH3. What do you think we will do to get the moles of NH3?
Theres some formuala...
Nah. You have 24 mol which equals to 3 parts. Find for one part and multiple by 2 to get the mols of NH3
Uhm...One part meaning?
Can you show me, I feel so lost...
Sure. So 24 mol is equal to 3 parts of H2 we use based on the equation so we divide 24 by 3. Then we know we get 2 parts of NH3 from the equation so we multiply it by 2
24/3*2 Get it? Want me to show it where there's whiteboard and mic?
so would it be 4 as the final answer?
Yeah. 4 mols. Do you understand what we did though?
NOOOOOO. 24/3 is 8 multiplied by 2 is 16
You know there is 24 mol. In 3 equal parts oh H2 and 2 equal parts of NH3?
Where do you get that information though? and oh okay just error in what comes first! sorry
Right. This is why we always need the balanced equation for stoichiometry.
Can you show me on the whiteboard?
Sure. Can you make a meeting on twiddla.com ? I can't because I already made one. Sent the link to me in PM
Actually its okay! i think i understood! Thanks so much for the help
Alright good job. Have a good night