Ammonium sulfate, an important chemical fertilizer, can be prepared by the reaction of ammonia with sulfuric acid according to the following balanced equation:

Question

2 NH3(g) + H2SO4(aq) → (NH4)2SO4(aq)

If a reaction vessel has 5.1 L of NH3 at 41.8°C and 25.6 atm, how many grams of H2SO4 are needed to completely react with it?

_________________ g H2SO4. Do NOT enter unit. Report your final answer with 3 SFs.
TBH i don't know how to do it. IF someone can help I'll Appreciate it! and Thank you! :)

Answer 1

I assume the 5.1 L NH3 is as a gas. So how many moles is that?
Use PV = nRT.
You know P = 25.5 atm
You know V = 5.1 L
n = ?
R = 0.0821 L*atm/mol*K
T = 41.8 C = 273 + 41.8 = ? K . Use K in the above equation.
Since the equation tells you 2 mol NH3 = 1 mol H2SO4, the mols H2SO4 = 1/2 mols NH3.
grams H2SO4 = mols H2SO4 x molr mass H2SO4 = ?
Post your work if you get stuck.

Answer 2

oops. Typo.
P = 25.6 atm.

Answer 3

N= PT v1/RT1 =
(25.6) (5.1) / (0.0821) (314.6) =
130.56 / 25.82866 = 5.0548
(5.0548) (

I don't know what I am doing LMAO

Answer 4

So far you're doing great, except for a typo. You copied the T twice and you meant n = P1V1/RT1. (N= PT v1/RT1 The T on the left is a typo). Also I think you meant 314.8 for K but that's not much of a problem. I obtained 5.05 as a rounded number. So
n = 5.0548 mols is essentially orrect. Now just do the rest.
You take half of that to get mols H2SO4 then multiply by molar mass of H2SO4.
I can see you're terribly confused. After you finish the problem tell me what is confusing you and why you don't think you know what you're doing.