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nikdorinn [45]
3 years ago
8

Which substance is the reducing agent in this reaction? 2KMnO4+3Na2SO3+H2O→2MnO2+3Na2SO4+2KOH

Chemistry
2 answers:
olganol [36]3 years ago
6 0
I believe <span>Na2SO3  is the solution to the problem.</span>
viva [34]3 years ago
6 0
The reducing agent would be the substance that is oxidized in the reaction or the one who loses electrons. From the reaction given, it would be KMnO4 that is  the reducing agent. It loses electrons From (MnO4)2- to MnO2. Hope this answers the question.
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Answer:

65.08 g.

Explanation:

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<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>

2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.

  • Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:

<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>

<u><em>Using cross multiplication:</em></u>

2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.

0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.

∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃  = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.

<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol ) = <em>65.08 g.</em>

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Which process would require the least amount of energy?
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