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Nutka1998 [239]
2 years ago
6

Practice: Determine the products, balance the

Chemistry
1 answer:
Anna11 [10]2 years ago
4 0

1. Inorganic combustion

2. Organic combustion

<h3>Further explanation</h3>

Given

The chemical equations

Required

Balanced equation

Solution

Organic compounds have the characteristic that there is a chain of carbon atoms, while inorganic does not have a chain of carbon atoms

The reaction of a substance with oxygen is called a combustion reaction

Complete combustion of carbon compounds will get CO₂ gas, whereas if it is not complete it will produce CO gas

So Organic : contain a carbon atom (and often a hydrogen atom)

1. Ca + O₂ ⇒ CaO₂

Inorganic combustion⇒does not contain carbon

2. C₃H₂O + 3O₂ → 3CO₂ + H₂O

Organic combustion

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If 9.85 grams of copper metal react with 31.0 grams of silver nitrate, how many grams of copper nitrate can be formed and how ma
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Answer:-

29.07 gram of Cu(NO3)2 will be formed.

4.756 grams of AgNO3 will be left over when the reaction is complete.

Explanation:-

Atomic weight of Cu = 63.546 g mol -1

Molecular weight of AgNO3 = 107.87 x 1 + 14 x 1 + 16 x 3

= 169.87 g mol-1

Number of moles of Copper = 9.85 gram / (63.546 g mol-1)

= 0.155 mol

Number of moles of AgNO3 = 31 gram / ( 169.87 g mol-1)

= 0.183 mol

The balanced chemical equation for this reaction is

Cu + AgNO3 --> Cu(NO3)2 + Ag

According to the equation,

1 mole of Cu reacts with 1 mole of AgNO3.

∴0.155 mol of Cu react with 0.155 mol of AgNO3.

Number of moles of AgNO3 left over = 0.183-0.155=0.028 mol

Number of grams of AgNO3 left over = 0.028 mol x 169.87 grams mol-1

= 4.756 gram

Molecular weight of Cu(NO3)2 = 63.546 x 1 + (14 x 1 +16 x 3 ) x 2

=187.546 gram

Now from the balanced chemical equation,

1 Cu gives 1 Cu(NO3)2

∴ 63.546 g of Cu gives 187.546 gram of Cu(NO3)2

9.85 grams pf Cu gives 187.546 x 9.85 / 64.546 gram of Cu(NO3)2

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Answer: The coefficients for the given reaction species are 1, 6, 2, 3.

Explanation:

The given reaction equation is as follows.

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Now, the two half-reactions can be written as follows.

Reduction half-reaction: Cr_{2}O^{2-}_{7} + 3e^{-} \rightarrow Cr^{3+}

This will be balanced as follows.

Cr_{2}O^{2-}_{7} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_{2}O ... (1)

Oxidation half-reaction: Cl^{-} \rightarrow Cl_{2} + 1e^{-}

This will be balanced as follows.

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Adding both equation (1) and (2) we will get the resulting equation as follows.

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