Answer:
K = 0,00000135 = 1.35 * 10^-6
Explanation:
Step 1: Data given
The equilibrium constant, K, for any reaction is defined as the concentrations of the products raised by their coefficients divided by the concentrations of the reactants raised by their coefficients. In this case, the concentrations are given as partial pressures.
The partial pressures of H2O = 0.0500 atm
The partial pressures of H2 = 0.00150 atm
The partial pressures of O2 = 0.00150 atm
Step 2: The balanced equation
2H2O(g) ⇆ 2H2(g) + O2(g)
Step 3: Calculate K
K = [O2][H2]² / [H2O]²
K = 0.00150 * 0.00150² / 0.0500²
K = 0,00000135 = 1.35 * 10^-6
Answer:
Therefore 9 moles of nitrogen are required to produce 18 moles of ammonia.
Answer:
81.8 %
Explanation:
The balanced equation for the reaction is
⇒ ![2 NH_3](https://tex.z-dn.net/?f=2%20NH_3)
So to find the number of moles of
we do 6.04kg ÷ 2 = 3.02 moles.
The mole ratio between
and
is 3:2. So to find the number of moles you do
× 2 which gives you 2.0124 moles.
Then you find the mass by doing moles multiplied by Mr: 2.0124 x 17 which gives you 34.21 kg.
Then to find the percentage yield you do:
x 100 = 81.82.......
Which is 81.8 %