Answer:
3Ba(OH)2 + 2H3PO4 —> Ba3(PO4)2 + 6H2O
Explanation:
Ba(OH)2 + H3PO4 —> Ba3(PO4)2 + H2O
There are 3 atoms of Ba on the right side and 1atom on the left side. It can be balance by putting 3 in front of Ba(OH)2 as shown below:
3Ba(OH)2 + H3PO4 —> Ba3(PO4)2 + H2O
There are 2 atoms of P on the right side and 1atom on the left. It can be balance by putting 2 in front of H3PO4 as shown below:
3Ba(OH)2 + 2H3PO4 —> Ba3(PO4)2 + H2O
Now, there are a total of 12 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 6 in front of H2O as shown below:
3Ba(OH)2 + 2H3PO4 —> Ba3(PO4)2 + 6H2O
Now the equation is balanced as the numbers of the atoms of the different elements present on both sides are equal
Answer:
At high temperatures or in the presence of catalysts, sulfur dioxide reacts with hydrogen sulfide to form elemental sulfur and water. This reaction is exploited in the Claus process, an important industrial method to dispose of hydrogen sulfide.
As we have the balanced reaction equation is:
N2O4 (g) ↔ 2NO2(g)
from this balanced equation, we can get the equilibrium constant expression
KC = [NO2]^2[N2O4]^1
from this expression, we can see that [NO2 ] is with 2 exponent of the stoichiometric and we can see that from the balanced equation as NO2
is 2NO2 in the balanced equation.
and [N2O4] is with 1 exponent of the stoichiometric and we can see that from the balanced equation as N2O4 is 1 N2O4 in the balanced equation.
∴ the correct exponent for N2O4 in the equilibrium constant expression is 1
Answer:
The new partial pressures after equilibrium is reestablished:
Explanation:
At equilibrium before adding chlorine gas:
Partial pressure of the 
Partial pressure of the 
Partial pressure of the 
The expression of an equilibrium constant is given by :
At equilibrium after adding chlorine gas:
Partial pressure of the 
Partial pressure of the 
Partial pressure of the 
Total pressure of the system = P = 263.0 Torr
At initail
(13.2) Torr (32.8) Torr (13.2) Torr
At equilbriumm
(13.2-x) Torr (32.8-x) Torr (217.0+x) Torr
Solving for x;
x = 6.402 Torr
The new partial pressures after equilibrium is reestablished:
