A certain liquid X has a normal boiling point of 133.60°C and a boiling point elevation constant Kb= 2.46°C kg mol^-1.Calculate
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Answer:
0.00369 moles of HCl react with carbonate.
Explanation:
Number of moles of HCl present initially = moles = 0.00600 moles
Neutralization reaction (back titration):
According to above equation, 1 mol of NaOH reacts with 1 mol of 1 mol of HCl.
So, excess number of moles of HCl present = number of NaOH added for back titration = moles = 0.00231 moles
So, mole of HCl reacts with carbonate = (Number of moles of HCl present initially) - (excess number of moles of HCl present) = (0.00600 - 0.00231) moles = 0.00369 moles
Hence, 0.00369 moles of HCl react with carbonate.
Answer:- 3.12 g carbon tetrachloride are needed.
Solution:- The balanced equation is:
From given actual yield and percent yield we will calculate the theoretical yield that would be further used to calculate the grams of carbon tetrachloride.
percent yield formula is:
percent yield =
theoretical = 3.44 g
From balanced equation, there is 2:1 mol ratio between dichloethane and carbon tetrachloride.
Molar mass of dichloroethane is 84.93 gram per mol and molar mass of carbon tetrachloride is 153.82 gram per mol.
=
So, 3.12 grams of carbon tetrachloride are needed to be reacted.
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