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3 years ago

Balanced equation of potassium chloride and ammonium nitrate; it’s types, net and observation

Chemistry

1 answer:

ASHA 777 [7]3 years ago

5 0

Potassium chloride reacts with ammonium nitrate to give ammonium chloride and potassium nitrate.

This is a type of double displacement reaction. The balanced chemical equation can be represented as,

$KCl(aq)+NH_{4}NO_{3}(aq)-->NH_{4}Cl(aq)+KNO_{3}(aq)$

Total ionic equation for this reaction will be,

$K^{+}(aq)+Cl^{-}(aq)+NH_{4}^{+}(aq)+NO_{3}^{-}(aq)--> K^{+}(aq)+NO_{3}^{-}(aq)+NH_{4}^{+}(aq)+Cl^{-}(aq)$

There is no apparent reaction as this reaction is not accompanied by the formation of a gas or a solid precipitate. We cannot observe any visual reaction as there is not net reaction taking place. All the ions remain as spectator ions.

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For each of the following balanced chemical equations, calculate how many moles and how many grams of each product would be prod

natima [27]

Answer:

(a)

Moles of ammonium chloride = 0.5 mole

Mass of ammonium chloride formed = 26.7455 g

(b)

Mole of $CS_2$ = 0.125 mole

Mass = Moles * Molar mass = 0.125 * 76.139 g = 9.52 g

Mole of $H_2S$ = 0.25 mole

Mass = Moles * Molar mass = 0.25 * 34.1 g = 8.525 g

Explanation:

(a)

For the first reaction:-

$NH_3_{(g)}+HCl_{(g)}\rightarrow NH_4Cl_{(s)}$

The mole ratio of the reactants = 1 : 1

0.5 moles of ammonia react with 0.5 moles of hydrochloric gas to give 0.5 moles of ammonium chloride

So, Moles of ammonium chloride formed = 0.5 moles

Molar mass of ammonium chloride = 53.491 g/mol

Mass = Moles * Molar mass = 0.5 * 53.491 g = 26.7455 g

(b)

For the first reaction:-

$CH_4_{(g)}+4S_{(s)}\rightarrow CS_2_{(l)}+2H_2S_{(g)}$

The mole ratio of the reactants = 1 : 4

It means

0.5 moles of methane react with 2.0 moles of sulfur to give 0.5 moles of Carbon disulfide and 1.0 moles of hydrogen sulfide gas.

But available moles of S = 0.5 moles

Limiting reagent is the one which is present in small amount. Thus, S is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

4 moles of S produces 1 mole of $CS_2$

Thus,

0.5 moles of S produces $\frac{1}{4}\times 0.5$ mole of $CS_2$

Mole of $CS_2$ = 0.125 mole

Molar mass of $CS_2$ = 76.139 g/mol

Mass = Moles * Molar mass = 0.125 * 76.139 g = 9.52 g

4 moles of S produces 2 moles of $H_2S$

Thus,

0.5 moles of S produces $\frac{2}{4}\times 0.5$ mole of $H_2S$

Mole of $H_2S$ = 0.25 mole

Molar mass of $H_2S$ = 34.1 g/mol

Mass = Moles * Molar mass = 0.25 * 34.1 g = 8.525 g

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Answer:

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Explanation:

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Kaylis [27]

Answer:

see explaination

Explanation:

Molecular equation;

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Total ionic equation; . Includes all ions ;

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3 years ago