Question

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balancing equations worksheet
balance the equations below using the spaces provided

1. n₂ + o₂ → __n₂o
2. co₂ + h₂o → c₆h₁₂o₆ + o₂
3. zn + hcl → zncl₂ + h₂
4. sicl₄ + h₂o → h₄sio₄ + hcl
5. al(oh)₃ + h₂so₄ → al₂(so₄)₃ + h₂o
6. fe₂(so₄)₃ + koh → k₂so₄ + fe(oh)₃
7. h₂so₃ + hi → h₂s + i₂ + __h₂o
8. fes₂ + o₂ → fe₂o₃ + so₂

balancing equations (advanced!)

1. sicl₄ + h₂o → sio₂ + hcl
2. as + naoh → na₃aso₃ + h₂
3. as₂s₃ + h₂ → as + h₂s
4. + hcl → cucl₂ + h₂o

Explanation:

Let's take the first equation: $\boldsymbol{\ce{N_{2} + O_{2} -> N_{2}O}}$ (the original might have a typo, assuming it's $\ce{N_{2}O}$ or $\ce{NO}$; let's correct to balancing $\ce{N_{2} + O_{2} -> N_{2}O}$ first, but maybe it's $\ce{2N_{2} + O_{2} -> 2N_{2}O}$? Wait, no, let's do the first visible one properly. Wait, the first equation is $\boldsymbol{\ce{N_{2} + O_{2} -> N_{2}O}}$ (assuming the user's first equation is $\ce{N_{2} + O_{2} -> N_{2}O}$). Let's balance it step by step.

Step 1: List atoms on each side

• Left (Reactants): $\ce{N}$: 2, $\ce{O}$: 2
• Right (Products): $\ce{N}$: 2, $\ce{O}$: 1

Step 2: Balance Oxygen

To balance $\ce{O}$, multiply $\ce{N_{2}O}$ by 2 (so $\ce{O}$ becomes 2). Now:

• Right: $\ce{N}$: $2 \times 2 = 4$, $\ce{O}$: $2 \times 1 = 2$

Step 3: Balance Nitrogen

Now $\ce{N}$ on left is 2, right is 4. Multiply $\ce{N_{2}}$ by 2:

• Left: $\ce{N}$: $2 \times 2 = 4$, $\ce{O}$: 2

Step 4: Final Balanced Equation

$\boldsymbol{\ce{2N_{2} + O_{2} -> 2N_{2}O}}$

Wait, but maybe the intended product is $\ce{NO}$ (common typo). Let's check that case: $\ce{N_{2} + O_{2} -> NO}$

• Left: $\ce{N}$: 2, $\ce{O}$: 2
• Right: $\ce{N}$: 1, $\ce{O}$: 1

Balance $\ce{N}$: multiply $\ce{NO}$ by 2 → $\ce{NO_2}$? No, $\ce{NO}$: multiply $\ce{NO}$ by 2 (right: $\ce{N}$:2, $\ce{O}$:2). Then equation: $\boldsymbol{\ce{N_{2} + O_{2} -> 2NO}}$

Let's take another equation, like $\boldsymbol{\ce{Zn + HCl -> ZnCl_{2} + H_{2}}}$ (third equation, labeled "3)").

Step 1: List atoms

• Left: $\ce{Zn}$: 1, $\ce{H}$: 1, $\ce{Cl}$: 1
• Right: $\ce{Zn}$: 1, $\ce{H}$: 2, $\ce{Cl}$: 2

Step 2: Balance $\ce{Cl}$ and $\ce{H}$

Multiply $\ce{HCl}$ by 2 (so $\ce{Cl}$: 2, $\ce{H}$: 2). Now:

• Left: $\ce{Zn}$: 1, $\ce{H}$: 2, $\ce{Cl}$: 2
• Right: $\ce{Zn}$: 1, $\ce{H}$: 2, $\ce{Cl}$: 2

Step 3: Final Equation

$\boldsymbol{\ce{Zn + 2HCl -> ZnCl_{2} + H_{2}}}$

For the equation $\boldsymbol{\ce{SiCl_{4} + H_{2}O -> H_{2}SiO_{3} + HCl}}$ (fourth equation, "4)"):

Step 1: List atoms

• Left: $\ce{Si}$: 1, $\ce{Cl}$: 4, $\ce{H}$: 2, $\ce{O}$: 1
• Right: $\ce{Si}$: 1, $\ce{Cl}$: 1, $\ce{H}$: $2 + 1 = 3$ (wait, $\ce{H_{2}SiO_{3}}$ has 2 $\ce{H}$, $\ce{HCl}$ has 1 $\ce{H}$ → total $\ce{H}$: 3? No, $\ce{H_{2}SiO_{3}}$: 2 $\ce{H}$, $\ce{HCl}$: 1 $\ce{H}$ per molecule. Let's re-express:

• Left: $\ce{Si}$:1, $\ce{Cl}$:4, $\ce{H}$:2 (from $\ce{H_{2}O}$), $\ce{O}$:1 (from $\ce{H_{2}O}$)
• Right: $\ce{Si}$:1 (from $\ce{H_{2}SiO_{3}}$), $\ce{Cl}$:1 (from $\ce{HCl}$), $\ce{H}$: $2 + 1 = 3$ (from $\ce{H_{2}SiO_{3}}$ and $\ce{HCl}$), $\ce{O}$:3 (from $\ce{H_{2}SiO_{3}}$)

Step 2: Balance $\ce{Cl}$

Multiply $\ce{HCl}$ by 4 (so $\ce{Cl}$: 4). Now:

• Right: $\ce{Cl}$: 4, $\ce{H}$: $2 + 4 = 6$ (from $\ce{H_{2}SiO_{3}}$ and $\ce{4HCl}$), $\ce{O}$:3

Step 3: Balance $\ce{H}$ and $\ce{O}$

$\ce{H}$ on left: 2 (from $\ce{H_{2}O}$). To get 6 $\ce{H}$ on left, multiply $\ce{H_{2}O}$ by 3 (so $\ce{H}$: $3 \times 2 = 6$, $\ce{O}$: $3 \times 1 = 3$). Now:

• Left: $\ce{Si}$:1, $\ce{Cl}$:4, $\ce{H}$:6, $\ce{O}$:3
• Right: $\ce{Si}$:1, $\ce{Cl}$:4, $\ce{H}$:6, $\ce{O}$:3

Step 4: Final Balanced Equation

$\boldsymbol{\ce{SiCl_{4} + 3H_{2}O -> H_{2}SiO_{3} + 4HCl}}$

Example 2: $\boldsymbol{\ce{Zn + HCl -> ZnCl_{2} + H_{2}}}$ (third equation)

• Step 1: Atoms: $\ce{Zn}$:1, $\ce{H}$:1, $\ce{Cl}$:1 (left); $\ce{Zn}$:1, $\ce{H}$:[LLM SSE On Failure]

Answer:

Let's take the first equation: $\boldsymbol{\ce{N_{2} + O_{2} -> N_{2}O}}$ (the original might have a typo, assuming it's $\ce{N_{2}O}$ or $\ce{NO}$; let's correct to balancing $\ce{N_{2} + O_{2} -> N_{2}O}$ first, but maybe it's $\ce{2N_{2} + O_{2} -> 2N_{2}O}$? Wait, no, let's do the first visible one properly. Wait, the first equation is $\boldsymbol{\ce{N_{2} + O_{2} -> N_{2}O}}$ (assuming the user's first equation is $\ce{N_{2} + O_{2} -> N_{2}O}$). Let's balance it step by step.

Step 1: List atoms on each side

• Left (Reactants): $\ce{N}$: 2, $\ce{O}$: 2
• Right (Products): $\ce{N}$: 2, $\ce{O}$: 1

Step 2: Balance Oxygen

To balance $\ce{O}$, multiply $\ce{N_{2}O}$ by 2 (so $\ce{O}$ becomes 2). Now:

• Right: $\ce{N}$: $2 \times 2 = 4$, $\ce{O}$: $2 \times 1 = 2$

Step 3: Balance Nitrogen

Now $\ce{N}$ on left is 2, right is 4. Multiply $\ce{N_{2}}$ by 2:

• Left: $\ce{N}$: $2 \times 2 = 4$, $\ce{O}$: 2

Step 4: Final Balanced Equation

$\boldsymbol{\ce{2N_{2} + O_{2} -> 2N_{2}O}}$

Wait, but maybe the intended product is $\ce{NO}$ (common typo). Let's check that case: $\ce{N_{2} + O_{2} -> NO}$

• Left: $\ce{N}$: 2, $\ce{O}$: 2
• Right: $\ce{N}$: 1, $\ce{O}$: 1

Balance $\ce{N}$: multiply $\ce{NO}$ by 2 → $\ce{NO_2}$? No, $\ce{NO}$: multiply $\ce{NO}$ by 2 (right: $\ce{N}$:2, $\ce{O}$:2). Then equation: $\boldsymbol{\ce{N_{2} + O_{2} -> 2NO}}$

Let's take another equation, like $\boldsymbol{\ce{Zn + HCl -> ZnCl_{2} + H_{2}}}$ (third equation, labeled "3)").

Step 1: List atoms

• Left: $\ce{Zn}$: 1, $\ce{H}$: 1, $\ce{Cl}$: 1
• Right: $\ce{Zn}$: 1, $\ce{H}$: 2, $\ce{Cl}$: 2

Step 2: Balance $\ce{Cl}$ and $\ce{H}$

Multiply $\ce{HCl}$ by 2 (so $\ce{Cl}$: 2, $\ce{H}$: 2). Now:

• Left: $\ce{Zn}$: 1, $\ce{H}$: 2, $\ce{Cl}$: 2
• Right: $\ce{Zn}$: 1, $\ce{H}$: 2, $\ce{Cl}$: 2

Step 3: Final Equation

$\boldsymbol{\ce{Zn + 2HCl -> ZnCl_{2} + H_{2}}}$

For the equation $\boldsymbol{\ce{SiCl_{4} + H_{2}O -> H_{2}SiO_{3} + HCl}}$ (fourth equation, "4)"):

Step 1: List atoms

• Left: $\ce{Si}$: 1, $\ce{Cl}$: 4, $\ce{H}$: 2, $\ce{O}$: 1
• Right: $\ce{Si}$: 1, $\ce{Cl}$: 1, $\ce{H}$: $2 + 1 = 3$ (wait, $\ce{H_{2}SiO_{3}}$ has 2 $\ce{H}$, $\ce{HCl}$ has 1 $\ce{H}$ → total $\ce{H}$: 3? No, $\ce{H_{2}SiO_{3}}$: 2 $\ce{H}$, $\ce{HCl}$: 1 $\ce{H}$ per molecule. Let's re-express:

• Left: $\ce{Si}$:1, $\ce{Cl}$:4, $\ce{H}$:2 (from $\ce{H_{2}O}$), $\ce{O}$:1 (from $\ce{H_{2}O}$)
• Right: $\ce{Si}$:1 (from $\ce{H_{2}SiO_{3}}$), $\ce{Cl}$:1 (from $\ce{HCl}$), $\ce{H}$: $2 + 1 = 3$ (from $\ce{H_{2}SiO_{3}}$ and $\ce{HCl}$), $\ce{O}$:3 (from $\ce{H_{2}SiO_{3}}$)

Step 2: Balance $\ce{Cl}$

Multiply $\ce{HCl}$ by 4 (so $\ce{Cl}$: 4). Now:

• Right: $\ce{Cl}$: 4, $\ce{H}$: $2 + 4 = 6$ (from $\ce{H_{2}SiO_{3}}$ and $\ce{4HCl}$), $\ce{O}$:3

Step 3: Balance $\ce{H}$ and $\ce{O}$

$\ce{H}$ on left: 2 (from $\ce{H_{2}O}$). To get 6 $\ce{H}$ on left, multiply $\ce{H_{2}O}$ by 3 (so $\ce{H}$: $3 \times 2 = 6$, $\ce{O}$: $3 \times 1 = 3$). Now:

• Left: $\ce{Si}$:1, $\ce{Cl}$:4, $\ce{H}$:6, $\ce{O}$:3
• Right: $\ce{Si}$:1, $\ce{Cl}$:4, $\ce{H}$:6, $\ce{O}$:3

Step 4: Final Balanced Equation

$\boldsymbol{\ce{SiCl_{4} + 3H_{2}O -> H_{2}SiO_{3} + 4HCl}}$

Example 2: $\boldsymbol{\ce{Zn + HCl -> ZnCl_{2} + H_{2}}}$ (third equation)

• Step 1: Atoms: $\ce{Zn}$:1, $\ce{H}$:1, $\ce{Cl}$:1 (left); $\ce{Zn}$:1, $\ce{H}$:[LLM SSE On Failure]