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Dima020 [189]
3 years ago
9

Suppose 50.0g of silver nitrate is reacted with 50g of hydrochloric acid producing silver chloride and a mixture of other produc

ts. If the mass of the other products formed is 46.4 grams, how much silver chloride was produced?
Chemistry
1 answer:
docker41 [41]3 years ago
8 0

Answer:

53.6 grams of silver chloride was produced.

Explanation:

AgNO_3+HCl+\rightarrow AgCl+HNO_3

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

Mass of silver nitrate = 50.0 g

Mass of hydrogen chloride = 50.0 g

Mass of silver chloride = x

Mass of  nitric acid = 46.4 g

Mass of silver nitrate + Mass of hydrogen chloride =

                             Mass of silver chloride + Mass of  nitric acid

[te]50.0 g+50.0 g=x+46.4 g[/tex]

x=50.0 g+50.0 g - 46.4 g = 53.6 g

53.6 grams of silver chloride was produced.

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Answer:

The ballance half reactions are:

Mg²⁺  + 2e⁻ → Mg

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Explanation:

Si (s) + Mg(OH)₂ (s) → Mg (s) + SiO₃²⁻ (aq)

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Let's take a look to half reactions:

Mg²⁺  + 2e⁻ → Mg

Si  → SiO₃²⁻ + 4e⁻

In basic medium, we have to add water, as the same amount of oxygen we have, IN THE SAME SIDE. We have 3 oxygens in products, so we add 3 H₂O and in the opposite site we can add OH⁻, to balance the hydrogen. The half reaciton will be:

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

If we want to ballance the main reaction we have to multiply (x2) the half reaction of oxidation. So the electrons can be ballanced.

2Mg²⁺  + 4e⁻ → 2Mg

Now, that they are ballanced we can sum the half reactions:

2Mg²⁺  + 4e⁻ → 2Mg

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

2Mg²⁺  + 4e⁻  + 6OH⁻ + Si  → 2Mg  +  SiO₃²⁻ + 4e⁻ + 3 H₂O

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