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Fantom [35]
3 years ago
14

How many liters of 4.0 M NaOH solution will react with 1.2 mol H2SO4? (Remember to balance the equation.)

Chemistry
2 answers:
valina [46]3 years ago
8 0
The answer to this is D
nydimaria [60]3 years ago
6 0

Answer:

The answer is D. 0.60 L

Explanation:

The balanced reaction equation including states of matter is;

H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)

More simple:

H2SO4 + 2NaOH → Na2SO4 + 2H2O

Now, we can see from this reaction equation that the mole ratio of NaOH to H2SO4 is 2:1

Number of moles of H2SO4 reacted = 1.2 moles

Hence;

2 moles of NaOH reacts with 1 mole of H2SO4

x moles of NaOH reacts with 1.2 moles of H2SO4

x = 2 * 1.2/1 = 2.4 moles of NaOH

Recall that;

Number of moles = Concentration * Volume

Volume = number of moles/concentration

Volume of NaOH  is obtained from;

Volume = 2.4 moles/ 4.0 M

Volume =  0.60 L

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<u>Answer:</u> S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction

<u>Explanation:</u>

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when oxidation number of a species increases.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when oxidation number of a species decreases.

For the given chemical reaction:

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<u>On the reactant side:</u>

Oxidation number of H = +1

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<u>On the product side:</u>

Oxidation number of H = +1

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As the oxidation number of S is increasing from 0 to +6. Thus, it is getting oxidized. Similarly, the oxidation number of N is decreasing from +5 to +2. Thus, it is getting reduced.

The oxidation numbers of O and H remain the same on both sides of the reaction. Thus, they are neither getting oxidized or reduced.

Hence, S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction

3 0
2 years ago
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