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1 year ago

59 Q.No. 28 b Hydrogen peroxide can act as oxidizing as well as reducing agent with example.

Chemistry

1 answer:

professor190 [17]1 year ago

3 0

Answer:

Hydrogen peroxide can function as an oxidizing agent as well as reducing agent.

H2O2 act as oxidizing agent in acidic medium.

Explanation:

Example : 2FeSO4 +H2SO4 +H2O2 —>

(ferrous sulphate)

Fe2(SO4)3 +2H2O

(ferric sulphate)

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A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

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59 Q.No. 28 b Hydrogen peroxide can act as oxidizing as well as reducing agent with example. ​

59 Q.No. 28 b Hydrogen peroxide can act as oxidizing as well as reducing agent with example.