Answer:
Explanation:
CH₃CHOHCOOH ⇄ CH₃CHOHCOO⁻ + H⁺
ionisation constant = 1.36 x 10⁻⁴ .
molecular weight of lactic acid = 90 g
moles of acid used = 20 / 90
= .2222
it is dissolved in one litre so molar concentration of lactic acid formed
C = .2222M
Let n be the fraction of moles ionised
CH₃CHOHCOOH ⇄ CH₃CHOHCOO⁻ + H⁺
C - nC nC nC
By definition of ionisation constant Ka
Ka = nC x nC / C - nC
= n²C ( neglecting n in the denominator )
n² x .2222 = 1.36 x 10⁻⁴
n = 2.47 x 10⁻²
nC = 2.47 x 10⁻² x .2222
= 5.5 x 10⁻³
So concentration of hydrogen or hydronium ion = 5.5 x 10⁻³ g ion per litre .
Answer:
Theoretical yield of HI is 512 g.
The percent yield for this reaction is 25%.
Explanation:

Moles of hydrogen gas = 3.0 moles
Moles of iodine gas = 2.0 moles
According to reaction 1 mol of hydrogen gas reacts with 1 mol of iodine gas.
Then 3.0 moles of hydrogen gas reacts with 3.0 mol of iodine gas. But there are 2.0 moles of iodine gas. Hence,Iodine is a limiting reagent. The production of HI will depend upon iodine gas moles.
According to reaction , 1 mol of iodine gas gives 2 moles of HI.
Then 2 moles of iodine gas will give:
of HI
Theoretically we will get 4 moles of HI.
Theoretical yield of HI = 4 mol × 128 g/mol= 512 g
Experimental yield of HI = 1.0 mol
= 1 mol × 128 g/mol= 128 g


The percent yield for this reaction is 25%.
Answer: 64.8
Explanation:Please see attachment for explanation
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