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Nastasia [14]
3 years ago
11

HNO3 + S --> H2SO4 + NO Now identify the element oxidized and the element reduced. Which element is oxidized? Which element i

s reduced? Which elements undergo no oxidation or reduction?
Chemistry
1 answer:
OleMash [197]3 years ago
3 0

<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:

HNO_3+S\rightarrow H_2SO_4+NO

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

Oxidation number of H = +1

Oxidation number of N = +5

Oxidation number of O = -2

Oxidation number of S = 0

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

Oxidation number of H = +1

Oxidation number of N = +2

Oxidation number of O = -2

Oxidation number of S = +6

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

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When methane ( CH4 ) burns, it reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reac
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<h2>Answer:</h2>1.33*10^{-2}grams

<h2>Explanations</h2>

The complete balanced equation for the given reaction is expressed as;

CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)

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\begin{gathered} moles\text{ of CH}_4=\frac{mass}{molar\text{ mass}} \\ moles\text{ of CH}_4=\frac{0.0059}{16.04} \\ moles\text{ of CH}_4=0.000368moles \end{gathered}

According to stoichimetry, 1 mole of methane produces 2 moles of water, hence the moles of water required will be:

\begin{gathered} moles\text{ of H}_2O=\frac{2}{1}\times0.000368 \\ moles\text{ of H}_2O=0.000736moles \end{gathered}

Determine the mass of water produced

\begin{gathered} Mass\text{ of H}_2O=moles\times molar\text{ mass} \\ Mass\text{ of H}_2O=0.000736\times18.02 \\ Mass\text{ of H}_2O=0.0133grams=1.33\times10^{-2}grams \end{gathered}

Therefore the mass of water produced from the complete combustion of 5.90×10−3 g of methane is 1.33 * 10^-2grams

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1 year ago
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Answer:

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Explanation:

Hope this helps!

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The diagram below represents reaction of reactant A and reactant B combining to form product C.
OleMash [197]

Answer:B

Explanation:

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