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kotykmax [81]
3 years ago
6

Balancing Chemical Equations

Chemistry
1 answer:
Citrus2011 [14]3 years ago
6 0

Start by assigning the most complicated species a coefficient of one. That species should contain the greatest number of elements, e.g. \text{HCl}.

\text{Zn} + \text{HCl} \to 1\; \text{ZnCl}_2 + \text{H}_2 <em>(not balanced)</em>

Assign coefficients to the rest of the species based on the conservation of atoms. For instance, the left hand side of the equation now contains one atom of hydrogen H and one atoms of chlorine Cl. The left hand side shall have an identical configuration. Both zinc chloride \text{ZnCl}_2 and hydrogen \text{H}_2 should therefore have a coefficient of 1/2. (Don't panic about the fractions. They are to be eliminated in a few more steps.)

\text{Zn} + 1\; \text{HCl} \to 1/2 \; \text{ZnCl}_2 + 1/2 \; \text{H}_2 <em>(not balanced)</em>

The coefficient 1/2 in front of zinc chloride indicates the presence of 1/2 zinc atom in the right hand side of the equation. Zinc on the left hand side of the equation should accordingly have a coefficient of 1/2.

1/2 \; \text{Zn} + 1\; \text{HCl} \to 1/2 \; \text{ZnCl}_2 + 1/2 \; \text{H}_2

Increasing coefficients on both sides of the equation by a factor of two to eliminate all fractions. Hence the balanced equation.

1 \; \text{Zn} + 2 \; \text{HCl} \to 1 \; \text{ZnCl}_2 + 1 \; \text{H}_2 (balanced)

The same set of operations should work for the second equation.

4 \; \text{Fe} + 3\; \text{O}_2 \to 2\; \text{Fe}_2\text{O}_3 (balanced)

Note that the third equation does not accurately represent the catalytic decomposition of hydrogen peroxide \text{H}_2\text{O}_2. The balanced equation should be:

2\; \text{H}_2\text{O}_2 \to 2\; \text{H}_2\text{O} + \text{O}_2 (balanced)


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1) Variables

  • molarity: M
  • density of the solution: d
  • moles of NaCl: n₁
  • mass of NaCl: m₁
  • molar mass of NaCl: MM₁
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    ⇒ mass in grams = number of moles × molar mass

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4) Number of moles of NaCl: n₁

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   M = n₁ / Vt

   ⇒ n₁ = M × Vt


5) Substitue in the equation of m₁:

   m₁ = M × Vt × MM₁


6) Substitute in the equation of density:

    d = [M × Vt × MM₁ + m₂] / (1000Vt)


7) Simplify and solve for M

  • d = M × Vt × MM₁ / (1000Vt) + m₂/ (1000Vt)
  • d = M × MM₁ / (1000) + m₂/ (1000Vt)

Making the simplistic assumption that the dissolved NaCl(s) does not affect the volume of the solvent water means 1000Vt = V₂  

  • d = M × MM₁ / (1000) + m₂/ V₂

        m₂/ V₂ is the density of water: 1.00 g/mL

  • d = M × MM₁ / (1000) + 1.00 g/mL
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  • M = [1,000/58.5] d - [1,000/ 58.5]

Comparing with the equation Molarity = m×density + b, you obtain:

  • m = 1,000/58.5
  • b = - 1,000/58.5
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