As chemistry aqa urgent help needed or i will suffer!

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Maxistan YT
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#1
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#1
Hi, can someone help me with this question, please?

Nitrogen monoxide reacts with chlorine to form nitrosyl chloride (NOCL).

2NO(g)+Cl2(g) = 2NOCl(g)

1.5 mol of NO are mixed with 1.00 mol of Cl2 and the mixture is left to reach equilibrium at a given temperature.
The equilibrium mixture contains 0.350 mol of NOCl

Calculate the amount, in moles, of NO and of Cl2 in the equilibrium mixture. (2 marks)

Give the expression for the equilibrium constant, Kc, for the reaction between nitrogen monoxide and chlorine to form nitrosyl chloride. (1 mark)

A different equilibrium mixture is prepared in a flask of volume 800 cm3 at a different temperature.
At equilibrium this mixture contains 0.850 mol of NO and 0.458 mol of Cl2. for the reaction at this temperature Kc = 1.32*10-2 mol-1dm3
Determine the amount in moles of NOCl in this equilibrium mixture. (4 marks)

Thank you.
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charco
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#2
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#2
(Original post by Maxistan YT)
Hi, can someone help me with this question, please?

Nitrogen monoxide reacts with chlorine to form nitrosyl chloride (NOCL).

2NO(g)+Cl2(g) = 2NOCl(g)

1.5 mol of NO are mixed with 1.00 mol of Cl2 and the mixture is left to reach equilibrium at a given temperature.
The equilibrium mixture contains 0.350 mol of NOCl

Calculate the amount, in moles, of NO and of Cl2 in the equilibrium mixture. (2 marks)

Give the expression for the equilibrium constant, Kc, for the reaction between nitrogen monoxide and chlorine to form nitrosyl chloride. (1 mark)

A different equilibrium mixture is prepared in a flask of volume 800 cm3 at a different temperature.
At equilibrium this mixture contains 0.850 mol of NO and 0.458 mol of Cl2. for the reaction at this temperature Kc = 1.32*10-2 mol-1dm3
Determine the amount in moles of NOCl in this equilibrium mixture. (4 marks)

Thank you.
I'll help you if you tell me(us) what the problem is.

Nitrogen monoxide reacts with chlorine to form nitrosyl chloride (NOCL).

2NO(g)+Cl2(g) = 2NOCl(g)

1.5 mol of NO are mixed with 1.00 mol of Cl2 and the mixture is left to reach equilibrium at a given temperature.
The equilibrium mixture contains 0.350 mol of NOCl

Calculate the amount, in moles, of NO and of Cl2 in the equilibrium mixture.

The equation tells you the amount of NOCl formed in the reaction. If you know the amount reacting then you can work out the amounrt formed, and vice versa.

How many moles of NO have to react to form 0.350 mol of NOCl?
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Maxistan YT
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#3
Report Thread starter 1 year ago
#3
(Original post by charco)
I'll help you if you tell me(us) what the problem is.

Nitrogen monoxide reacts with chlorine to form nitrosyl chloride (NOCL).

2NO(g)+Cl2(g) = 2NOCl(g)

1.5 mol of NO are mixed with 1.00 mol of Cl2 and the mixture is left to reach equilibrium at a given temperature.
The equilibrium mixture contains 0.350 mol of NOCl

Calculate the amount, in moles, of NO and of Cl2 in the equilibrium mixture.

The equation tells you the amount of NOCl formed in the reaction. If you know the amount reacting then you can work out the amounrt formed, and vice versa.

How many moles of NO have to react to form 0.350 mol of NOCl?
This is my method:

So the KC expression for the reaction is:

(NOCL)^2 / (NO)^2 x (CL2) , kc= 1.32x10^-2 NO= 0.85 mol CL2= 0.458 mol Volume = 800cm^3 ( This should be converted to 0.8 dm^3)

The question asks for us to find the (NOCL)^2 :

So NOCL^2 = KC x (NO)^2 x (CL2)

This is (1.32*10^-2 ) x (0.85/0.8)^2 x (0.458/0.8 ) = 8.53x10^-3

Now NOCL^2 = 8.53 x 10^-3 , we square root to find it to be 0.09236419507

Now i multiply 0.09223 number by 0.8 which is 0.074
If i divide the 0.09223 number by 0.8 it's 0.115455

IDK if i did it correct, can you tell me what i did wrong ?
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anon25x
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(Original post by Maxistan YT)
This is my method:

So the KC expression for the reaction is:

(NOCL)^2 / (NO)^2 x (CL2) , kc= 1.32x10^-2 NO= 0.85 mol CL2= 0.458 mol Volume = 800cm^3 ( This should be converted to 0.8 dm^3)

The question asks for us to find the (NOCL)^2 :

So NOCL^2 = KC x (NO)^2 x (CL2)

This is (1.32*10^-2 ) x (0.85/0.8)^2 x (0.458/0.8 ) = 8.53x10^-3

Now NOCL^2 = 8.53 x 10^-3 , we square root to find it to be 0.09236419507

Now i multiply 0.09223 number by 0.8 which is 0.074
If i divide the 0.09223 number by 0.8 it's 0.115455

IDK if i did it correct, can you tell me what i did wrong ?
I got 1.125 and 0.825 is that correct?
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Merryl69_0
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#5
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(Original post by anon25x)
I got 1.125 and 0.825 is that correct?
Yh
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deskochan
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#6
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#6
(Original post by Maxistan YT)
This is my method:

So the KC expression for the reaction is:

(NOCL)^2 / (NO)^2 x (CL2) , kc= 1.32x10^-2 NO= 0.85 mol CL2= 0.458 mol Volume = 800cm^3 ( This should be converted to 0.8 dm^3)

The question asks for us to find the (NOCL)^2 :

So NOCL^2 = KC x (NO)^2 x (CL2)

This is (1.32*10^-2 ) x (0.85/0.8)^2 x (0.458/0.8 ) = 8.53x10^-3

Now NOCL^2 = 8.53 x 10^-3 , we square root to find it to be 0.09236419507

Now i multiply 0.09223 number by 0.8 which is 0.074
If i divide the 0.09223 number by 0.8 it's 0.115455

IDK if i did it correct, can you tell me what i did wrong ?
Now I multiply the 0.09223 number by 0.8 which is 0.074.
This is correct because the concentration x volume = moles
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