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

Number of moles of carbonate

The ions left in solution are Na^+ and NO3^-

Number of moles of calcium nitrate  = 100/1000 L × 1 = 0.1 moles

Since;

1 mole of sodium carbonate reacts with 1 mole of calcium nitrate  then 0.1 moles of sodium carbonate were used.

Conductivity of filtrate

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

The calculations are performed as follows

  • 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

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

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.

Learn more about MnO4^- : https://brainly.com/question/10887629