In which laboratory process could a student use 0.10 M NaOH(aq) to determine the concentration of an aqueous solution of HBr?
2 answers:
Answer:
4) titration
Explanation:
Titration is a standard process used in a laboratory to determine the concentration of an unknown analyte. A titrant of known concentration is gradually added to a known volume of the analyte in the presence of a suitable indicator. The end of the titration is marked by a color change of the analyte.
The given example is that of an acid(HBr) - base(NaOH) titration which can be represented by the following equation:
NaOH + HBr → NaBr + H2O
Thus 1 mole of acid gets neutralized by 1 mole of the base to form 1 mole of the salt (NaBr)
Let M1 and V1 are the molarity and volume of the base (NaOH). Here, the molarity of NaOH is known = M1 = 0.10 M and the volume, V1 corresponds to the end point in the titration.
M2 and V2 are the molarity and volume of HBr. Here, V2 is known whereas M2 needs to be determined.
Based on the reaction stoichiometry:
moles of NaOH = moles of HBr
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Explanation:
The given data is as follows.
Mass flow rate of mixture = 1368 kg/hr
in feed = 40 mole%
This means that in feed = (100 - 40)% = 60%
We assume that there are 100 total moles/hr of gas in feed stream.
Hence, calculate the total mass flow rate as follows.
40 moles/hr of N_{2}/hr (28 g/mol of ) + 60 moles/hr of
(2 g/mol of
)
= 1120 g/hr + 120 g/hr
= 1240 g/hr
= (as 1 kg = 1000 g)
= 1.240 kg/hr
Now, we will calculate mol/hr in the actual feed stream as follows.
= 110322.58 moles/hr
It is given that amount of nitrogen present in the feed stream is 40%. Hence, calculate the flow of into the reactor as follows.
= 44129.03 mol/hr
As 1 mole of nitrogen has 28 g/mol of mass or 0.028 kg.
Therefore, calculate the rate flow of into the reactor as follows.
= 1235.612 kg/hr
Thus, we can conclude that the the feed rate of pure nitrogen to the mixer is 1235.612 kg/hr.