Chemistry worksheet for students to practice identifying and balancing chemical reactions, featuring various chemical equations.

Let’s go through each reaction one by one. For each, we’ll:

1. Balance the equation (make sure atoms on left = atoms on right).
2. Identify the type of reaction (synthesis, decomposition, single replacement, double replacement, or combustion).

We’ll do this step by step for all 13 reactions.

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(i) Na₂CO₃ + HCl → NaCl + H₂CO₃

First, balance it:
- Left: Na=2, C=1, O=3, H=1, Cl=1
- Right: Na=1, Cl=1, H=2, C=1, O=3

Need 2 NaCl and 2 HCl to balance Na and Cl and H.

→ Balanced: Na₂CO₃ + 2HCl → 2NaCl + H₂CO₃

Type: Two compounds swap parts → Double Replacement

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(ii) NH₄NO₂ → N₂ + H₂O

Left: N=2, H=4, O=2
Right: N=2, H=2, O=1 → need 2 H₂O

→ Balanced: NH₄NO₂ → N₂ + 2H₂O

Type: One compound breaks into two → Decomposition

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(iii) N₂ + O₂ → N₂O₅

Left: N=2, O=2
Right: N=2, O=5 → need to make O even. Try 2N₂O₅ → O=10, so need 5O₂

Then N: 4 on right → need 2N₂ on left

→ Balanced: 2N₂ + 5O₂ → 2N₂O₅

Type: Two elements combine → Synthesis

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(iv) MgCO₃ → MgO + CO₂

Already balanced?
Left: Mg=1, C=1, O=3
Right: Mg=1, C=1, O=1+2=3 → yes!

→ Balanced: MgCO₃ → MgO + CO₂

Type: One compound breaks → Decomposition

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(v) KBr + Cl₂ → KCl + Br₂

Left: K=1, Br=1, Cl=2
Right: K=1, Cl=1, Br=2 → need 2KBr and 2KCl

→ Balanced: 2KBr + Cl₂ → 2KCl + Br₂

Type: Element replaces another in compound → Single Replacement

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(vi) Zn + CuSO₄ → Cu + ZnSO₄

Check atoms:
Zn=1, Cu=1, S=1, O=4 on both sides → already balanced!

→ Balanced: Zn + CuSO₄ → Cu + ZnSO₄

Type: Element replaces another → Single Replacement

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(vii) P + O₂ → P₄O₆

Left: P=1, O=2
Right: P=4, O=6 → need 4P and 3O₂

→ Balanced: 4P + 3O₂ → P₄O₆

Type: Elements combine → Synthesis

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(viii) SrBr₂ + (NH₄)₂CO₃ → SrCO₃ + NH₄Br

Left: Sr=1, Br=2, N=2, H=8, C=1, O=3
Right: Sr=1, C=1, O=3, N=1, H=4, Br=1 → need 2 NH₄Br

→ Balanced: SrBr₂ + (NH₄)₂CO₃ → SrCO₃ + 2NH₄Br

Type: Swap ions → Double Replacement

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(ix) AgNO₃ + (NH₄)₂CrO₄ → Ag₂CrO₄ + NH₄NO₃

Left: Ag=1, NO₃=1, NH₄=2, CrO₄=1
Right: Ag=2, CrO₄=1, NH₄=1, NO₃=1 → need 2AgNO₃ and 2NH₄NO₃

→ Balanced: 2AgNO₃ + (NH₄)₂CrO₄ → Ag₂CrO₄ + 2NH₄NO₃

Type: Swap ions → Double Replacement

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(x) K + H₂O → KOH + H₂

Left: K=1, H=2, O=1
Right: K=1, O=1, H=1+2=3 → too many H on right.

Try 2K + 2H₂O → 2KOH + H₂

Check:
Left: K=2, H=4, O=2
Right: K=2, O=2, H=2+2=4 → good!

→ Balanced: 2K + 2H₂O → 2KOH + H₂

Type: Element replaces H in water → Single Replacement

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(xi) Al + Pb₂(NO₃)₂ → Al(NO₃)₃ + Pb

Left: Al=1, Pb=2, N=2, O=6
Right: Al=1, N=3, O=9, Pb=1 → not balanced.

Find LCM for NO₃: 6 and 3 → use 6.

So: 2Al(NO₃)₃ → needs 6 NO₃ → so 3 Pb₂(NO₃)₂ → gives 6 NO₃ and 6 Pb

Then Al: 2 on right → 2 on left
Pb: 6 on left → 6 on right

→ Balanced: 2Al + 3Pb₂(NO₃)₂ → 2Al(NO₃)₃ + 6Pb

Wait — that’s messy. Let me check again.

Actually, Pb₂(NO₃)₂ has 2 Pb and 2 NO₃.

Al(NO₃)₃ has 1 Al and 3 NO₃.

To balance NO₃: LCM of 2 and 3 is 6.

So: 3 Pb₂(NO₃)₂ → 6 NO₃ and 6 Pb
2 Al(NO₃)₃ → 6 NO₃ and 2 Al

So Al: 2 on left
Pb: 6 on right → so 6 Pb on right

→ Balanced: 2Al + 3Pb₂(NO₃)₂ → 2Al(NO₃)₃ + 6Pb

Type: Element replaces metal → Single Replacement

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(xii) Fe + O₂ → Fe₃O₄

Left: Fe=1, O=2
Right: Fe=3, O=4 → need 3Fe and 2O₂? 2O₂=4O → yes.

→ Balanced: 3Fe + 2O₂ → Fe₃O₄

Type: Elements combine → Synthesis

Wait — actually, Fe₃O₄ is magnetite, and this is a combination reaction. Yes, synthesis.

But note: sometimes this is called “combustion” if it’s burning, but since no hydrocarbon, better to call it synthesis.

Actually, standard classification: metal + oxygen → metal oxide = Synthesis

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Now let’s list all answers clearly.

Final Answer:

(i) Balanced: Na₂CO₃ + 2HCl → 2NaCl + H₂CO₃; Type: Double Replacement
(ii) Balanced: NH₄NO₂ → N₂ + 2H₂O; Type: Decomposition
(iii) Balanced: 2N₂ + 5O₂ → 2N₂O₅; Type: Synthesis
(iv) Balanced: MgCO₃ → MgO + CO₂; Type: Decomposition
(v) Balanced: 2KBr + Cl₂ → 2KCl + Br₂; Type: Single Replacement
(vi) Balanced: Zn + CuSO₄ → Cu + ZnSO₄; Type: Single Replacement
(vii) Balanced: 4P + 3O₂ → P₄O₆; Type: Synthesis
(viii) Balanced: SrBr₂ + (NH₄)₂CO₃ → SrCO₃ + 2NH₄Br; Type: Double Replacement
(ix) Balanced: 2AgNO₃ + (NH₄)₂CrO₄ → Ag₂CrO₄ + 2NH₄NO₃; Type: Double Replacement
(x) Balanced: 2K + 2H₂O → 2KOH + H₂; Type: Single Replacement
(xi) Balanced: 2Al + 3Pb₂(NO₃)₂ → 2Al(NO₃)₃ + 6Pb; Type: Single Replacement
(xii) Balanced: 3Fe + 2O₂ → Fe₃O₄; Type: Synthesis

Note: There are only 12 items listed in the image (i to xii), though labeled up to (xii). The original says "(xiii)" at end but it's likely typo — we did 12.

Wait — looking back: user wrote up to (xii) Fe + O₂ → Fe₃O₄, and then said "(xiii)" but didn’t write anything. Probably mistake. We’ll stop at (xii).

Actually, counting the Roman numerals in the image description: i, ii, iii, iv, v, vi, vii, viii, ix, x, xi, xii — that’s 12.

So final answer includes 12 reactions.

Let me write them neatly.

Final Answer:
(i) Na₂CO₃ + 2HCl → 2NaCl + H₂CO₃ ; Double Replacement
(ii) NH₄NO₂ → N₂ + 2H₂O ; Decomposition
(iii) 2N₂ + 5O₂ → 2N₂O₅ ; Synthesis
(iv) MgCO₃ → MgO + CO₂ ; Decomposition
(v) 2KBr + Cl₂ → 2KCl + Br₂ ; Single Replacement
(vi) Zn + CuSO₄ → Cu + ZnSO₄ ; Single Replacement
(vii) 4P + 3O₂ → P₄O₆ ; Synthesis
(viii) SrBr₂ + (NH₄)₂CO₃ → SrCO₃ + 2NH₄Br ; Double Replacement
(ix) 2AgNO₃ + (NH₄)₂CrO₄ → Ag₂CrO₄ + 2NH₄NO₃ ; Double Replacement
(x) 2K + 2H₂O → 2KOH + H₂ ; Single Replacement
(xi) 2Al + 3Pb₂(NO₃)₂ → 2Al(NO₃)₃ + 6Pb ; Single Replacement
(xii) 3Fe + 2O₂ → Fe₃O₄ ; Synthesis