Answer:
Kc = 3.72 × 10⁶
Explanation:
Let's consider the following reaction:
NH₄HS(g) ⇄ NH₃(g) + H₂S(g)
At equilibrium, we have the following concentrations:
[NH₄HS] = 0.196 M (assuming a 1 L flask)
[NH₃] = 9.56 × 10² M
[H₂S] = 7.62 × 10² M
We can replace this data in the Kc expression.
![Kc=\frac{[NH_{3}] \times [H_{2}S] }{[NH_{4}HS]} =\frac{9.56 \times 10^{2} \times 7.62 \times 10^{2}}{0.196} =3.72 \times 10^{6}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BNH_%7B3%7D%5D%20%5Ctimes%20%5BH_%7B2%7DS%5D%20%7D%7B%5BNH_%7B4%7DHS%5D%7D%20%3D%5Cfrac%7B9.56%20%5Ctimes%2010%5E%7B2%7D%20%20%5Ctimes%207.62%20%20%5Ctimes%2010%5E%7B2%7D%7D%7B0.196%7D%20%3D3.72%20%5Ctimes%2010%5E%7B6%7D)
As we can see the chemical equation is balanced.K3PO4 + Al(NO3)3 → 3KNO3 + AlPO4
So, by principle of conservation of mass when 1 mole of K3PO4 reacts with 1 mol of Al(NO3)3 , it peoduces 3 mol of KNO3 and 1 mol of AlPO4
So, when 2.5 moles of potassium phosphate react and Al(NO3)3 is present in excess , 2.5*3= 7.5 mol of KNO3 is formed
Answer:
see explanation below
Explanation:
First to all, this is a redox reaction, and the reaction taking place is the following:
2KMnO4 + 3H2SO4 + 5H2O2 -----> 2MnSO4 + K2SO4 + 8H2O + 5O2
According to this reaction, we can see that the mole ratio between the peroxide and the permangante is 5:2. Therefore, if the titration required 21.3 mL to reach the equivalence point, then, the moles would be:
MhVh = MpVp
h would be the hydrogen peroxide, and p the permanganate.
But like it was stated before, the mole ratio is 5:2 so:
5MhVh = 2MpVp
Replacing moles:
5nh = 2MpVp
Now, we just have to replace the given data:
nh = 2MpVp/5
nh = 2 * 1.68 * 0.0213 / 5
nh = 0.0143 moles
Now to get the mass, we just need the molecular mass of the peroxide:
MM = 2*1 + 2*16 = 34 g/mol
Finally the mass:
m = 0.0143 * 34
m = 0.4862 g