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3 years ago

Balancing Chemical Equations

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. $\text{HCl}$ .

$\text{Zn} + \text{HCl} \to 1\; \text{ZnCl}_2 + \text{H}_2$ (not balanced)

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.)

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

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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What is the balanced form of the following equation? Br 2 + S 2 O 3 2– + H 2 O → Br 1– + SO 4 2– + H +

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Answer:

4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻

Explanation:

Br₂ + S₂O₃²⁻ + H₂O → Br⁻ + SO₄²⁻ + H⁺

This is a redox reaction:

Br₂ changes the oxidation state from 0 to -1, so it was reduced

In the S₂O₃⁻² anion S changes the oxidation state from +2 to +6 in sulfate anion. (S₂O₃⁻², it is called thiosulfate)

We have protons in the main equation, so we assume we are in acidic medium:

Br₂ + 2e⁻ → 2Br⁻ Reduction

We balanced the bromide with 2, so the bromine has gained 2 electrons.

5H₂O + S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8e- Oxidation

First of all, we add 2 to the sulfate anion in the product side, in order to balance the S.

As we have 8 O in right side, and 3 O in left side, we must add 5 O. We add 5 water in the place where the O are lower (reactant side).

Now, we have 10 H, in the reactant side, so we balance the product side with protons (10 H⁺).

Sulfur changed the oxidation state from +2 to +6, so it released 4 electrons, but, if you see thiosulfate anion you have 2 sulfurs so finally it has released 8 electrons.

Electrons are unbalanced so we multiply reduction x4, and oxidation x1.

(Br₂ + 2e⁻ → 2Br⁻) . 4 = 4Br₂ + 8e⁻ → 8Br⁻

(5H₂O + S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8e-) . 1 = STAYS THE SAME.

We sum both half reactions, to cancel the elecetrons:

4Br₂ + 8e⁻ + 5H₂O + S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8e- + 8Br⁻

Finally the balanced reaction is: 4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻

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4 years ago

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A. As pressure on the gas increases, the volume and temperature will both decrease

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4 years ago

5. 16.3 g of NaCl is dissolved in water to make 1.75 L of solution. What is the molarity of this solution? A 0.159 M B 0.278 M C

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Answer: The molarity of this solution is 0.159 M.

Explanation:

Given: Mass of solute = 16.3 g

Volume = 1.75 L

Number of moles is defined as the mass of substance divided by its molar mass.

Hence, moles of NaCl (molar mass = 58.44 g/mol) ar calculated as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{16.3 g}{58.44 g/mol}\\= 0.278 mol$

Molarity is the number of moles of a substance present in a liter of solution.

So, molarity of the given solution is calculated as follows.

$Molarity = \frac{no. of moles}{Volume (in L)}\\= \frac{0.278 mol}{1.75}\\= 0.159 M$

Thus, we can conclude that the molarity of this solution is 0.159 M.

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Answer:

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