Chemistry: Writing and Balancing Equations Homework online ... - Free Printable

I'll solve each problem step by step, carefully balancing the chemical equations.

Problem 1: Mg + O₂ → MgO

Let me balance this equation:
- Left side: Mg = 1, O = 2
- Right side: Mg = 1, O = 1

To balance oxygen, I need 2 MgO on the right:
Mg + O₂ → 2MgO

Now magnesium is unbalanced (1 on left, 2 on right), so I need 2 Mg on the left:
2Mg + O₂ → 2MgO

Checking: Mg = 2, O = 2 on both sides ✓

The coefficient on oxygen gas (O₂) is 1.

Answer: A

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Problem 2: F₂ + S₈ → S₂F₁₀

Let me balance this:
- Left side: F = 2, S = 8
- Right side: S = 2, F = 10

First, let's balance sulfur. I have 8 S on left and 2 S on right, so I need 4 S₂F₁₀:
F₂ + S₈ → 4S₂F₁₀

Now check fluorine: Left has 2 F, right has 4 × 10 = 40 F
So I need 20 F₂ on the left:
20F₂ + S → 4S₂F₁₀

Checking: F = 40, S = 8 on both sides ✓

The coefficient on fluorine gas (F₂) is 20.

Answer: D

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Problem 3: O₂ + N₂ → N₃O₅

Let me balance this:
- Left side: O = 2, N = 2
- Right side: N = 3, O = 5

I need to find coefficients that work. Let me try different combinations.

If I put 2 in front of N₃O₅:
O₂ + N₂ → 2N₃O₅

Right side: N = 6, O = 10

For nitrogen: I need 3 N₂ on left (3 × 2 = 6)
For oxygen: I need 5 O₂ on left (5 × 2 = 10)

5O₂ + 3N₂ → 2N₃O₅

Checking: O = 10, N = 6 on both sides ✓

The coefficient on nitrogen gas (N₂) is 3.

Answer: B

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Problem 4: K + H₂O → KOH + H₂

Let me balance this:
- Left side: K = 1, H = 2, O = 1
- Right side: K = 1, O = 1, H = 1 + 2 = 3

Hydrogen is unbalanced. Let me try putting 2 in front of H₂O:
K + 2H₂O → KOH + H₂

Left: K = 1, H = 4, O = 2
Right: K = 1, O = 1, H = 3

Still not balanced. Let me try 2KOH on the right:
K + 2H₂O → 2KOH + H₂

Left: K = 1, H = 4, O = 2
Right: K = 2, O = 2, H = 2 + 2 = 4

Now potassium is unbalanced. I need 2K on the left:
2K + 2H₂O → 2KOH + H₂

Checking: K = 2, H = 4, O = 2 on both sides ✓

The coefficient on potassium hydroxide (KOH) is 2.

Answer: B

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Problem 5: Balance the following equations from scratch

5a: __Al + __O₂ → __Al₂O

Left: Al, O₂
Right: Al₂O₃ (2 Al, 3 O)

To balance oxygen: I need even number of O on right. Try 2Al₂O₃ (which gives 6 O atoms)
__Al + __O₂ → 2Al₂O₃

For 6 O atoms on right, I need 3 O₂ on left:
__Al + 3O₂ → 2Al₂O

For aluminum: 2 × 2 = 4 Al on right, so I need 4 Al on left:
4Al + 3O₂ → 2Al₂O₃

Checking: Al = 4, O = 6 on both sides ✓

Answer: 4, 3, 2

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5b: __CO₂ + __H₂O → __C₁₂H₂₂O₁₁ + __O₂

This looks like photosynthesis reverse or sugar formation.

Right side: C₁₂H₂₂O₁₁ has 12 C, 22 H, 11 O

For carbon: Need 12 CO₂
For hydrogen: Need 11 H₂O (to get 22 H)

12CO₂ + 11H₂O → C₁₂H₂₂O₁₁ + __O₂

Let's count oxygen:
Left: 12 × 2 + 11 × 1 = 24 + 11 = 35 O
Right in sugar: 11 O
Remaining O for O₂: 35 - 11 = 24 O atoms = 12 O₂ molecules

12CO₂ + 11H₂O → C₁₂H₂₂O₁₁ + 12O₂

Checking: C = 12, H = 22, O = 35 on both sides ✓

Answer: 12, 11, 1, 12

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5c: __HNO₃ + __Cd → __H₂ + __Cd(NO₃)₂

Right side: Cd(NO₃)₂ has 1 Cd, 2 NO₃ groups

For nitrate: Need 2 HNO₃ to provide 2 NO₃ groups
2HNO₃ + __Cd → __H₂ + Cd(NO₃)₂

For cadmium: 1 Cd on each side
2HNO₃ + Cd → __H₂ + Cd(NO₃)₂

For hydrogen: 2 H on left, need 1 H₂ on right
2HNO₃ + Cd → H₂ + Cd(NO₃)₂

Checking: H = 2, N = 2, O = 6, Cd = 1 on both sides ✓

Answer: 2, 1, 1, 1

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5d: __Cu + __H₂O + __CO₂ + __O₂ → __Cu(OH)₂ + __CuCO₃

This is copper corrosion forming basic copper carbonate.

Let me think about this systematically. The products contain Cu(OH)₂ and CuCO₃.

Let's say we have 'a' Cu(OH)₂ and 'b' CuCO₃ on the right.

From Cu(OH)₂: needs Cu, 2 OH (which comes from H₂O)
From CuCO₃: needs Cu, CO₃ (which comes from CO₂ and possibly O₂)

Let me try: 2Cu + H₂O + CO₂ + O₂ → Cu(OH)₂ + CuCO₃

Checking atoms:
Left: Cu = 2, H = 2, O = 1 + 2 + 2 = 5, C = 1
Right: Cu = 2, H = 2, O = 2 + 3 = 5, C = 1

Perfect!

Answer: 2, 1, 1, 1, 1, 1

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5e: __CuCO₃ → __CuO + __CO₂

This is thermal decomposition of copper carbonate.

Left: CuCO (1 Cu, 1 C, 3 O)
Right: CuO + CO₂ (1 Cu, 1 C, 1 + 2 = 3 O)

Already balanced!

Answer: 1, 1, 1

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5f: __CH₁₀ + __O₂ → __CO₂ + __H₂O

This is butane combustion.

Left: C₄H₁₀ (4 C, 10 H)
Need: 4 CO₂ (for 4 C) and 5 H₂O (for 10 H)

C₄H₁₀ + __O₂ → 4CO₂ + 5H₂O

Counting oxygen on right: 4 × 2 + 5 × 1 = 8 + 5 = 13 O atoms
Need: 13/2 = 6.5 O₂

To avoid fractions, multiply everything by 2:
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

Checking: C = 8, H = 20, O = 26 on both sides ✓

Answer: 2, 13, 8, 10

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5g: __NaCl + __Sn(NO₃)₂ → __NaNO + __SnCl₂

This is a double displacement reaction.

Left: NaCl, Sn(NO₃)₂
Right: NaNO₃, SnCl₂

Sn(NO₃)₂ has 2 NO₃ groups, so I need 2 NaNO₃ on right.
That means I need 2 NaCl on left.

2NaCl + Sn(NO)₂ → 2NaNO₃ + SnCl₂

Checking: Na = 2, Cl = 2, Sn = 1, N = 2, O = 6 on both sides ✓

Answer: 2, 1, 2, 1

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5h: __Zn + __O₂ → __ZnO

Left: Zn, O₂
Right: ZnO (1 Zn, 1 O)

For oxygen: O₂ has 2 O atoms, so I need 2 ZnO on right
__Zn + O₂ → 2ZnO

For zinc: Need 2 Zn on left
2Zn + O₂ → 2ZnO

Checking: Zn = 2, O = 2 on both sides ✓

Answer: 2, 1, 2

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5i: __Fe + __HCl → __FeCl₃ + __H₂

Left: Fe, HCl
Right: FeCl₃ (1 Fe, 3 Cl), H₂

For chlorine: FeCl₃ has 3 Cl, so I need 3 HCl
__Fe + 3HCl → FeCl₃ + __H₂

For hydrogen: 3 H on left, but H₂ needs even number. Let me try 6 HCl:
__Fe + 6HCl → __FeCl₃ + __H₂

For FeCl₃: With 6 Cl from HCl, I can make 2 FeCl₃
__Fe + 6HCl → 2FeCl₃ + __H₂

For iron: Need 2 Fe on left
2Fe + 6HCl → 2FeCl₃ + __H₂

For hydrogen: 6 H on left = 3 H₂ on right
2Fe + 6HCl → 2FeCl₃ + 3H₂

Checking: Fe = 2, H = 6, Cl = 6 on both sides ✓

Answer: 2, 6, 2, 3

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Final Answer:

Multiple Choice:
1. A (1)
2. D (20)
3. B (3)
4. B (2)

Balanced Equations:
5a. 4Al + 3O₂ → 2Al₂O₃
5b. 12CO₂ + 11H₂O → C₁₂H₂₂O₁₁ + 12O₂
5c. 2HNO₃ + Cd → H₂ + Cd(NO₃)₂
5d. 2Cu + H₂O + CO₂ + O₂ → Cu(OH)₂ + CuCO₃
5e. CuCO₃ → CuO + CO₂
5f. 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
5g. 2NaCl + Sn(NO₃)₂ → 2NaNO₃ + SnCl₂
5h. 2Zn + O₂ → 2ZnO
5i. 2Fe + 6HCl → 2FeCl + 3H₂