Answer
% of oxygen(O) in glucose = 96/ 180 × 100 = 53.33
Explanation:
so the mass of oxygen in 1 mole of glucose = 6 × 16.00 g = 96.00 g. % oxygen = 96.00 × 100 % = 53.3 % by mass.
Answer:
62.07 %
Explanation:
Chemical Formula = (CH3)2CO = C3H6O
Mass of C = 12 g/mol
Mass of H = 1 g/mol
Mass of O = 12 g/mol
Mass of C3H6O = 3(12) + 6(1) + 16 = 58 g/mol
Total mass of C in Acetone = 12 * 3 = 36 g
Mass Percent of C = Total mass of C / Mass of Acetone * 100
Mass percent = 36 / 58 * 100
Mass percent = 62.07 %
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.