Question

2l of NH3 at 303k (30 degree Celsius) and 20.265 kpa is neutralised by 135 ml of sol of h2so4 .Find normality​

Answer 1

Explanation:

We need to get the moles of NH3:

PV = nRT

(0.20 atm) (2.0 L) = (n) (0.08206 L atm / mol K) (298 K)

n = 0.01636 mol

You did not give a temperature, so I assumed 25 C.

Now, the chemical equation for the neutralization:

2NH3 + H2SO4 ---> (NH4)2SO4

The key is that 2 NH3 are required for ever one H2SO4.

0.01636 mol / 2 = 0.00818 mol <--- that's how many moles of H2SO4 reacted

0.00818 mol / 0.134 L = 0.061 M (to two sig figs)

Edit: OK, now I'm upset. I missed the 30 C. Aargh!

(0.20 atm) (2.0 L) = (n) (0.08206 L atm / mol K) (303 K)

n = 0.016087 mol

0.016087 mol / 2 = 0.0080435 mol

0.0080435 mol / 0.134 L = 0.060 M

Fixed!

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