1 Hydrogen and chlorine react to produce hydrogen chloride (HCl) as shown below.
ID: 569474 • Letter: 1
Question
1 Hydrogen and chlorine react to produce hydrogen chloride (HCl) as shown below. At equilibrium, the flask contains 0.239 g of HCl, 0.254 g of Cl2 and 0.00013 g of H2. Calculate the value of Kc.
The reaction is: H2 + Cl2 <--> 2HCl
Hint: Work in mol/L.
1.72 x 103
5.4 x 10-2
1.84 x 102
3.55 x 10-5
2.99 x 104
2.If 40 g of HF are initially placed in an empty 1 L flask, calculate the amount of HF remaining if at equilibrium 1.0 g of H2 had been produced. The reaction is:
2HF --> H2 + F2
Hint: Work in moles and then switch back to grams at the end.
20 g of HF
25 g of HF
39 g of HF
40 g of HF
1.0 g of HF
1.72 x 103
5.4 x 10-2
1.84 x 102
3.55 x 10-5
2.99 x 104
2.If 40 g of HF are initially placed in an empty 1 L flask, calculate the amount of HF remaining if at equilibrium 1.0 g of H2 had been produced. The reaction is:
2HF --> H2 + F2
Hint: Work in moles and then switch back to grams at the end.
20 g of HF
25 g of HF
39 g of HF
40 g of HF
1.0 g of HF
Explanation / Answer
no of moles of HCl = W/G.M.Wt
= 0.239/36.5 = 0.00655moles
[HCl] = 0.00655mole/1 L = 0.00655mole/L
no of moles of Cl2 = W/G.M.Wt
= 0.254/71 = 0.00357 moles
[Cl2] = 0.00357mole/L = 0.00357mole/L
no of moles of H2 = W/G.M.Wt
= 0.00013/2 = 0.000065 moles
[H2] = 0.000065mole/1L = 0.000065mole/L
H2 + Cl2 <--> 2HCl
Kc = [HCl]^2/[H2][Cl2]
= (0.00655)^2/0.00357*0.000065
= 184
= 1.84*10^2 >>>>answer
2.
no of moles of HF = W/G.M.Wt
= 40/20 = 2 moles
no of moles of H2 = W/G.M.Wt
= 1/2 = 0.5 moles
2HF --> H2 + F2
I 2 0 0
C -2*0.5 0.5 0.5
E 1 0.5 0.5
no of moles of Hf at equilibrium = 1moles
mass of HF at equilibrium = 1*20 = 20g
20 g of HF >>>>answer
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