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Snezhnost [94]
3 years ago
12

AlCl3 + NaOH = NaClO2 + Al(OH)3 balance the equation! *dont forget the O2 in NaClO2!!

Chemistry
1 answer:
Verdich [7]3 years ago
6 0

Answer:

AlCl₃  + 3NaOH  → 3NaCl  +   Al(OH)₃

Explanation:

Problem is to balance the given chemical equation:

      Equation:  

              AlCl₃  + NaOH  → NaClO₂   +   Al(OH)₃

Balancing a chemical equation involves the conservation of atoms on both sides of the equation.

We can use a very simple mathematical method to balance the above equation;

           aAlCl₃  + bNaOH  → cNaCl   +   dAl(OH)₃

a, b, c and d are coefficients that will balance the equation:

  Conserving Al:  a  = d

                       Cl:   3a = c

                       Na:   b = c

                       O:      b = 3d

                       H:      b = 3d

Assume that a = 1;

                       c = 3

                       b = 3

                      d = 1

            AlCl₃  + 3NaOH  → 3NaCl  +   Al(OH)₃

O₂ in from of the NaClO₂ flouts the rule of chemical combination

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Will Upvote!
Dvinal [7]
If my memory serves me well, the correct answer is: Chemical processes store and release <span>C: energy</span><span> in the bonds of molecules.

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4 years ago
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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

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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

It is given that amount of nitrogen present in the feed stream is 40%. Hence, calculate the flow of N_{2} into the reactor as follows.

                       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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Answer:

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