The reaction to form NH3 is : N2 + 3H2-> 2NH3 12,33g NH3 is 12,33/17,03=0,3 =0,724 moles of NH3 moles NH3. So you need 1,5*0,724 = 1,086 moles H2 1,086*2,016 = 2,189 g of H2 is needed ro form 12,33 g NH3
What question? Lhh this is hilarious.
High temperature and pressure produce the highest rate of reaction. However, this must be balanced with the high cost of the energy needed to maintain these conditions. Catalysts increase the rate of reaction without affecting the yield. This can help create processes which work well even at lower temperatures.
I hope this helps you.
By using the ICE table :
initial 0.2 M 0 0
change -X + X +X
Equ (0.2 -X) X X
when Ka = (X) (X) / (0.2-X)
so by substitution:
4.9x10^-10 = X^2 / (0.2-X) by solving this equation for X
∴X ≈ 10^-6
∴[HCN] = 10^-6
and PH = -㏒[H+]
= -㏒ 10^-6
= 6
The answer would be 2.0 x 10^-1
