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nika2105 [10]
3 years ago
11

C5H12 + O2 = CO2 + H2O

Chemistry
1 answer:
Lady_Fox [76]3 years ago
3 0

Answer: 241.6 grams of CO2

Explanation: you take 84.3 grams C5H12 and divide it by 72.15 grams of C5H12(which is the molar mass) you take that answer and calculate the mols of CO2 by multiplying the 1.168 you got before and multiply it by 5. You take the answer you get from that and multiply it by the molar mass of CO2 and get the theoretical yield and then you just plug it in. 94= (x/257.02)x100 and solve to find x which is the actual yield.

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Pure nitrogen (N2) and pure hydrogen (H2) are fed to a mixer. The product stream has 40.0% mole nitrogen and the balance hydroge
LuckyWell [14K]

Explanation:

The given data is as follows.

        Mass flow rate of mixture = 1368 kg/hr

      N_{2} in feed = 40 mole%

This means that H_{2} in feed = (100 - 40)% = 60%

We assume that there are 100 total moles/hr of gas (N_{2} + H_{2}) in feed stream.

Hence, calculate the total mass flow rate as follows.

           40 moles/hr of N_{2}/hr (28 g/mol of N_{2}) + 60 moles/hr of H_{2}/hr (2 g/mol of H_{2})

                  40 \times 28 g/hr + 60 \times 2 g/hr    

                  = 1120 g/hr + 120 g/hr

                  = 1240 g/hr

                  = \frac{1240}{1000}              (as 1 kg = 1000 g)

                  = 1.240 kg/hr

Now, we will calculate mol/hr in the actual feed stream as follows.

                 \frac{100 mol/hr}{1.240 kg/hr} \times 1368 kg/hr

                   = 110322.58 moles/hr

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                       0.4 \times 110322.58 moles/hr

                      = 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 N_{2} into the reactor as follows.

                       0.028 kg \times 44129.03 mol/hr

                         = 1235.612 kg/hr

Thus, we can conclude that the the feed rate of pure nitrogen to the mixer is 1235.612 kg/hr.

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