i need some help on chemistry homework..

so possibly something like this?


Rb+ RbNO3---> Rb2O+N2

2 Rb 3?

3 O 3?
 
coolgirlspyer90 said:
so possibly something like this?


Rb+ RbNO3---> Rb2O+N2

2 Rb 3?

3 O 3?
Click to expand...

ok lets start with O if you have 3 on the reactant side you need 3 on the product side, but if you but 3 on the product side how does that affect Rb?
 
ember said:
ok lets start with O if you have 3 on the reactant side you need 3 on the product side, but if you but 3 on the product side how does that affect Rb?
Click to expand...

like, it would not become equal at all? and that i would need to change something..?
 
coolgirlspyer90 said:
oh and that it needs to be an equal number to all sides otherwise the equation would be unbalanced?
Click to expand...


now your wheels are turning keep playing with those numbers until they balance out I will give you a hint the first Rb will end up with a two digit number
 
ember said:
now your wheels are turning keep playing with those numbers until they balance out I will give you a hint the first Rb will end up with a two digit number
Click to expand...

so:

2 Rb 2-- which would equal 4

2 N 2-- equal 4 as well

3 O 1--equal 4??

thats balanced isn't it?
 
Rb+ RbNO3---> Rb2O+N2

I need 3 O so now I have 3 O on the product side but that gives me 6 Rb as well, dang it!!! Well how do I get those to match lets try another element how about N how would you balance those?
 
Balancing equations is harder because you can't really just plug and chug. You have to think about what steps are most intelligent. There's kind of a methodical pattern though.

For example, here's a step-by-step way to do the reaction ember was talking about.
We start with:

_Rb + _RbNO3 -> _Rb2O + _N2
There are 3 elements to be concerned with: Rb, N, and O. Right now, the status is:
Rb - 2 on reactant side, 2 on product side (balanced)
N - 1 on reactant side, 2 on product side (unbalanced)
O - 3 on reactant side, 1 on product side (unbalanced)

We can make adjustments by adding numbers to the _, which multiplies all the elements in the compound that it prefixes. A good rule of thumb is to try to make small adjustments first.

So let's try to balance the N. Replace the 2nd _ with a 2, the equation is then:
_Rb + 2RbNO3 -> _Rb20 + _N2
Rb - 3 on left, 2 on right (unbalanced)
N - 2 on left, 2 on right (balanced)
O - 6 on left, 1 on right (unbalanced)

To balance the Rb, we would have to add prefixes to both sides since 2 and 3 are relatively prime. So let's balance the O now because that's easier. We can do that by putting a 6 in the first _ on the product side. Rb20 has 1 O, so 6Rb20 will have 6 O's.
As a side effect, the Rb2 is also multiplied by 6, so now we have 12 Rb's on the right.

Now we have:
_Rb + 2RbNO3 -> 6Rb20 + _N2
Rb - 2 on left, 12 on right (unbalanced)
N - 2 on left, 2 on right (balanced)
O - 6 on left, 6 on right (balanced)

Since the first symbol on the reactant side is Rb alone, we can prefix it without affecting the amounts of other elements. So all we have to do is multiply it by 10 to increase the number of Rb's on the left by 10.

And we end up with:
10Rb + 2RbNO3 -> 6Rb20 + N2
(12 Rb's, 2 N's and 6 O's)

In short, write down the elements and how many of each are on each side. Then try to do the smallest possible change that will balance an additional element, and go from there. Sometimes you'll have to do larger changes and cross out stuff a lot, but this works for most simpler reactions.

See if you can apply this to the other problems. Hope that helps!
 
here are some examples of the mol stuff..hopefully u can understand how i did it

a. NaHCO3(s) --> NaOH(s) + CO2(g)
1(23)+1(1)+1(12)+3(16) --> 1(23)+1(16)+1(1)
=84 =40
given: 2.0 g NaHCO3
required: g NaOH
analysis: 84 g NaHCO3: 40 g NaOH
solution: 84 NaHCO3/40 NaOH= 2.0 NaHCO3/ x
84x=80
x= 0.95 g NaOH


b. 2NaHCO3(s) --> Na2CO3(s) + CO2(g) + H2O(g)
2(23)+2(1)+2(12)+6(16) --> 2(23)+1(12)+3(16)
=168 =106
given:2.0 g NaHCO3
required: g Na2CO3
analysis: 168 g NaHCO3: 106 g Na2CO3
solution 168 NaHCO3/ 106 Na2CO3= 2.0 NaHCO3/ x
168x=212
x= 1.3 g Na2CO3


c. 2NaHCO3(s) --> Na2O(s) + 2CO2(g) + H2O
2(23)+2(1)+2(12)+6(16) --> 2(23)+1(16)
=168 =62
given:2.0 NaHCO3
required: g Na2O
solution: 168 NaHCO3/ 62 Na2O= 2.0 HaHCO3/ x
168x=124
x=0.74 g Na2O

if u have any questions lemme know:)
 
ugh, chemistry...I had to take it a few years back, my teacher was from India, with an accent so thick that she could have stood in front of us going blah blah blah la la la la blah blah for 3 hours for all I could understand (maybe that is similar to what a deaf person experiences?)

Anyway, it was hopeless. Here I was with a subject I could barely comprehend in the first place and I wind up with a teacher that barely spoke my language. I went to Barnes & Noble and bought "Chemistry for Dummies" and taught myself chemistry that semester. Even got a B, which was a major shock in itself. Those Dummies books are awesome....maybe I should get ASL for Dummies?
 
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