# chemistry equilibrium help

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Thread starter 1 month ago
#1
Compounds A and B react together to form an equilibrium mixture containing compounds C and D according to the equation
2A + B ⇌ 3C + D
A beaker contained 40 cm3 of a 0.16 mol dm–3 aqueous solution of A.
9.5 × 10–3 mol of B and 2.8 × 10–2 mol of C were added to the beaker and the mixture was left to reach equilibrium.
The equilibrium mixture formed contained 3.9 × 10–3 mol of A.
Calculate the amounts, in moles, of B, C and D in the equilibrium mixture.

Initial amount of A = 6.4 × 10–3
Equ A = 6.4 × 10–3 - 2x so x=1.25x10-3

this is the part im stuck on; why can't you divide 6.4x10-3 by 2 to get x?
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1 month ago
#2
Because
equilibrium A = initial A -2x
=> 2x = initial A -equilibrium A
=> 2x = 6.4x10^-3. - 3.9x10^-3
=> 2x= 2.5x10^-3
=>. x = 1.25x10^-3

You have to work out the number moles of A that have reacted. So initial number of moles takeaway moles that are left in equilibrium = 2x
Last edited by GabiAbi84; 1 month ago
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Thread starter 1 month ago
#3
(Original post by GabiAbi84)
Because
equilibrium A = initial A -2x
=> 2x = initial A -equilibrium A
=> 2x = 6.4x10^-3. - 3.9x10^-3
=> 2x= 2.5x10^-3
=>. x = 1.25x10^-3

You have to work out the number moles of A that have reacted. So initial number of moles takeaway moles that are left in equilibrium = 2x
i dont understand why it's 2x
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1 month ago
#4
(Original post by onedance1)
i dont understand why it's 2x
From the formula you’re given the ratio of moles for each compound:

2A+B <=> 3C+D

So number of moles is x

2x of A + x of B <=> 3x of C + x of D

(I know those middle arrows aren’t the right sign, I don’t know how to type them on my phone)
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Thread starter 1 month ago
#5
(Original post by GabiAbi84)
From the formula you’re given the ratio of moles for each compound:

2A+B <=> 3C+D

So number of moles is x

2x of A + x of B <=> 3x of C + x of D

(I know those middle arrows aren’t the right sign, I don’t know how to type them on my phone)
ohh i see, i just dont get why i cant divide the number of moles of A by 2 cause that's what i do for every other calculation
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1 month ago
#6
(Original post by onedance1)
ohh i see, i just dont get why i cant divide the number of moles of A by 2 cause that's what i do for every other calculation
You are but you have to work out how many moles of A have been used first.
So you work out the intial number of moles
Then the question gives you the number of moles remaining at equilibrium
So you take away the number of moles left at equilibrium from the initial number of moles to get how many moles actually reacted.
Then divide that number by two for the equation
Last edited by GabiAbi84; 1 month ago
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