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NNADVOKAT [17]
3 years ago
7

Write balanced chemical equations for the following reactions:

Chemistry
2 answers:
vekshin13 years ago
8 0

Answer:

Solution given:

a) hydrogen gas + nitrogen gas gives ammonia.

<u>Balanced chemical equation:</u>

\boxed{\bold{\green{3H_{2}+N_{2}\rightarrow 2NH_{3}}}}

when <u>pure</u><u> </u><u>and</u><u> </u><u>dry</u><u> </u><u>Nitrogen</u><u> </u><u>and</u><u> </u><u>Hydrogen</u><u> </u><u>gas</u><u> </u><u>is</u><u> </u><u>passed</u><u> </u><u>in</u><u> </u><u>the</u><u> </u><u>ratio</u><u> </u><u>2</u><u>:</u><u>3</u><u> </u><u>at</u><u> </u><u>4</u><u>5</u><u>0</u><u>°</u><u>C</u><u> </u><u>temperature</u><u> </u><u>2</u><u>0</u><u>0</u><u>-</u><u>9</u><u>0</u><u>0</u><u> </u><u>ATM</u><u> </u><u>pressure</u><u> </u><u>in</u><u> </u><u>tye</u><u> </u><u>presence</u><u> </u><u>of</u><u> </u><u>iron</u><u> </u><u>as</u><u> </u><u>catalyst</u><u> </u><u>and</u><u> </u><u>molybdenum</u><u> </u><u>as</u><u> </u><u>promotor</u><u> </u><u>which</u><u> </u><u>forms</u><u> </u><u>Ammonia</u><u>.</u>

b) sodium peroxide + water gives sodium hydroxide + oxygen gas.

<u>Balanced chemical equation:</u>

\boxed{\green{\bold{3NaO_{2}+2H_{2}O \rightarrow 4NaOH +O_{2}}}}

<u>When sodium peroxide is combined with hot water double displacement reaction takes place and forms sodium hydroxide and oxygen gas along with heat.</u>

viktelen [127]3 years ago
6 0

Answer:

a) 3H₂ (g) + N₂ (g) → 2NH₃ (g)

b) 2Na₂O₂ + 2H₂O → 4NaOH + O₂

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Reactions rxn
  • Compounds

<u>Aqueous Solutions</u>

  • States of matter

Explanation:

a)

When we write this chemical reaction, we know that hydrogen and nitrogen are <em>diatomic</em> elements. Ammonia you just have to remember the chemical compound formula for. So our unbalanced equation would be:

H₂ (g) + N₂ (g) → NH₃ (g)

Now to balance this equation, we see that we have an uneven amount of hydrogens and nitrogens on both sides of the rxn. Let's balance out the nitrogens first by multiplying the products by 2:

H₂ (g) + N₂ (g) → 2NH₃ (g)

We see that now we have the number of nitrogens balanced on both sides, but our hydrogens are still unbalanced. Let's balance those by making the reactants the same number as our products:

  • We have 6 hydrogens now on the products side
  • 2H = 6H

It looks like we need to multiply 3 on the reactant hydrogens:

3H₂ (g) + N₂ (g) → 2NH₃ (g)

And we have our balanced formula!

b)

Same concept as A.

Recall how to write chemical compounds. The charge of sodium (Na) is +1 and the charge for polyatomic ion peroxide (O₂²⁻) is -2. Also recall the charge for polyatomic ion hydroxide (OH)m which is -1:

Sodium peroxide = Na₂O₂

Water = H₂O (standard knowledge)

Sodium hydroxide = NaOH

Oxygen gas = O₂

Write out our unbalanced rxn:

Na₂O₂ + H₂O → NaOH + O₂

Right away we can see that it is definitely unbalanced. We can see that we have an odd number of oxygens on both sides. We don't like odds here, so let's multiply 2 to the sodium peroxide and to make it even:

2Na₂O₂ + H₂O → NaOH + O₂

We can see that we have an even amount of oxygens on the reactant side. Now we have to balance the number of sodiums on the product side:

2Na₂O₂ + H₂O → 4NaOH + O₂

We now have the sodiums balanced. Moving onto the hydrogens. 2 on the reactant side and 4 on the product side:

2Na₂O₂ + 2H₂O → 4NaOH + O₂

We now have the hydrogens balanced. When we move on to oxygens, we can see that the number of oxygens have the same number of moles on both sides, and that would be our balanced rxn.

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