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Olegator [25]
3 years ago
11

AlCl3 + Na NaCl + Al Did Cl change oxidation number?

Chemistry
2 answers:
gizmo_the_mogwai [7]3 years ago
6 0
Yes. If this is the balanced equation:

AlCl3 + 3Na —— 3NaCl + Al

then Al was reduced from a 3+ oxidation (to neutralize the 3- from the chlorine) to a 0 oxidation (elemental ground state).
damaskus [11]3 years ago
5 0

Answer : There is no change takes place in the oxidation number of Cl.

Explanation :

The given chemical reaction is,

AlCl_3+Na\rightarrow NaCl+Al

This reaction is an unbalanced reaction because the chlorine atoms are not balanced.

In order to balance the chemical reaction, the coefficient 3 is put before the Na and NaCl.

The balanced chemical reaction will be,

AlCl_3+3Na\rightarrow 3NaCl+Al

Now we have to calculate the oxidation number of all the elements.

In AlCl_3, the oxidation number of Al and Cl are, (+3) and (-1) respectively.

The oxidation number Na and Al are, zero (0)

In NaCl, the oxidation number of Na and Cl are, (+1) and (-1) respectively.

From this we conclude that the oxidation number of Cl changes from (-1) to (-1) that means remains same.

Therefore, there is no change takes place in the oxidation number of Cl.

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A 0.00143 M  concentration of MnO4^- is not a reasonable solution .

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Since;

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The claim of the student that the concentration of sodium carbonate is too low is wrong because the value was calculated from concentration  and volume of calcium nitrate  and not using the precipitate. If the filtrate is tested for conductivity, it will be found to conduct electricity because it contains sodium and NO3 ions.

2) In the reaction as shown, the MnO4^- ion was reduced.

The initial volume is 3.4 mL while the final volume is 29.6 mL.

Number of moles of MnO4^- ion = (29.6 mL - 3.4 mL)/1000 × 0.0235 M = 0.0006157 moles

<h3>The calculations are performed as follows</h3>

  • If 2 moles of MnO4^- reacted with 5 moles of acid

0.0006157 moles of MnO4^- reacted with  0.0006157 moles ×  5 moles/ 2 moles

= 0.0015 moles

  • In this case, number of moles of acid = 0.139 g/90 g/mol = 0.0015 moles

Number of moles of MnO4^-  = 0.00143 M × (29.6 mL - 3.4 mL)/1000

= 0.000037 moles

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0.000037 moles of MnO4^- reacts with 0.000037 moles × 5 moles/ 2 moles

= 0.000093 moles

  • Hence, this is not a reasonable amount of solution.

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