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Rainbow [258]
3 years ago
10

Four moles of hydrogen chloride (HCl) gas react with one mole of oxygen (O2) gas to produce two moles of chlorine gas (Cl2) and

two moles of water gas (H2O). Using this description of the reaction, write the corresponding chemical equation, showing the appropriate coefficients for each reactant and product. (Include states-of-matter under the given conditions in your answer.)
Chemistry
2 answers:
Marianna [84]3 years ago
7 0

Answer:

4HCl(g) + O₂(g) →  2Cl₂(g) + 2H₂O(g)

Explanation:

In order to find the equation we should state:

The reactants → Hydrogen chloride (HCl) and oxygen (O₂)

The products → Chlorine gas (Cl₂) and water gas (H₂O)

The balanced equation is:

4HCl + O₂  →  2Cl₂ + 2H₂O

It is a redox reaction, where the oxygen reduces to make water, and the chloride is oxidized to produce elemental chlorine.

ANEK [815]3 years ago
7 0

Answer:

4HCl(g) + O2(g) → 2Cl2(g) + 2H2O(g)

Explanation:

Step 1: Data given

Number of moles hydrogen chloride (HCl) = 4.0 moles

Number of moles oxygen gas (O2) = 1.0 moles

Number of moles chlorine gas (Cl2) produced = 2.0 moles

Number of moles of water gas (H20) = 2.0 moles

Step 2: The unbalanced equation

HCl(g) + O2(g) → Cl2(g) + H2O(g)

Step 3: Balancing the equation

HCl(g) + O2(g) → Cl2(g) + H2O(g)

On the left side we have 2x O (in O2), on the right side we have 1x O (in H2O).

To balance the amount of O on both sides, we have to multiply H2O on the right side by 2.

HCl(g) + O2(g) → Cl2(g) + 2H2O(g)

On the left side we have 1x H (in HCl), on the right side we have 4x H (in 2H2O). To balance the amount of H on both sides, we have to multiply HCl on the left side by 4.

4HCl(g) + O2(g) → Cl2(g) + 2H2O(g)

On the left side we have 4x Cl (in 4HCl), on the right side we have 2x Cl (in Cl2). To balance the amount of Cl on both sides,we have to multiply Cl2 by 2. Now the equation is balanced.

4HCl(g) + O2(g) → 2Cl2(g) + 2H2O(g)

This means for 4 moles Hcl we need 1 mol O2 to react to produce 2 moles Cl2 and 2 moles H2O

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Answer : The radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.

Explanation :

As we are given that the Na⁺ radius is 56.4% of the Cl⁻ radius.

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So, the radius of Na⁺ = x\times \frac{56.4}{100}=(0.564x)pm

In the crystal structure of NaCl, 2 Cl⁻ ions present at the corner and 1 Na⁺ ion present at the edge of lattice.

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Given:

Distance between Na⁺ nuclei = 566 pm

Thus, the relation will be:

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