Posted by Jeena on .
SO2(g) + NO2(g) reverse reaction arrow SO3(g) + NO (g)
At a given temperature, analysis of an equilibrium mixture found [SO2] = 4.00 M, [NO2] = 0.500 M, [SO3] = 3.00 M, and [NO] = 2.00 M.
How many moles/liter of NO2 would have to be added to the original equilibrium mixture to increase the equilibrium concentration of SO3 from 3.00 M to 4.10 M at the same temperature?
How would you set this up? I tried to set it up using (4.1)(2.0+1.1)/(4.01.1)(.5+x) = 3, but got it wrong...and also tried (4.1x)(2.0+x)/(4.0x)(.5+X)=3...and also got it wrong...

Chemistry 
DrBob222,
Answered below. I can help find it if you can't.

Chemistry 
Jeena,
So I found that you would have to set it up using (4.1)(3.1)/(2.9)(x1.1). The x includes the .5 of NO2 already added, so to find how much was added to the original, find x and just subtract .5. The "x" value was the new initial value used to raise SO3 to 4.1 and that is why it is x1.1.

AS Maths 
jesse,
If you start at £18000, and you are given a 6% raise each year, how many years must pass before your salary is at least £180000?