Question

name period types of reactions worksheet given the reactions below, identify each as synthesis (s), decomposition (d), single replacement (sr), double replacement (dr), or combustion (c). afterwards, balance the reaction. 1) k + cl₂ → kcl type of reaction: __ 2) nh₃ → n₂ + h₂ type of reaction: ____ 3) naf + br₂ → nabr + f₂ type of reaction: __ 4) so₂ + li₂se → sse₂ + li₂o type of reaction: __ 5) pbso₄ → pbso₃ + o₂ type of reaction: ____ 6) al + fe₂o₃ → fe + al₂o₃ type of reaction: __ 7) sf₆ → s + f₂ type of reaction: ____ 8) c₂h₄ + o₂ → co₂ + h₂o type of reaction: __ 9) kclo₃ → kcl + o₂ type of reaction: ____ 10) pb(no₃)₂ + nacl → nano₃ + pbcl₂ type of reaction: __ 11) what type of reaction is shown to the left? (with image: + = , = oxygen, = carbon) 12) using the key provided, write the complete chemical formula shown in the box above: __ + __ → ____ 13) now balance the equation above so that both sides of the reaction are equal.

Explanation:

Step1: Identify Reaction Type (1)

The reaction is \( \text{K} + \text{Cl}_2 ightarrow \text{KCl} \). Synthesis (S) reactions combine substances. Here, K and \( \text{Cl}_2 \) form KCl.
To balance: Count atoms. Left: K=1, Cl=2; Right: K=1, Cl=1. Multiply KCl by 2: \( 2\text{KCl} \). Now K=2 on right, so multiply K by 2: \( 2\text{K} + \text{Cl}_2 ightarrow 2\text{KCl} \). Type: S.

Step2: Identify Reaction Type (2)

Reaction: \( \text{NH}_3 ightarrow \text{N}_2 + \text{H}_2 \). Decomposition (D) breaks a compound. Balance: Left N=1, H=3; Right N=2, H=2. Multiply \( \text{NH}_3 \) by 2: \( 2\text{NH}_3 \) (N=2, H=6). Then multiply \( \text{H}_2 \) by 3: \( 2\text{NH}_3 ightarrow \text{N}_2 + 3\text{H}_2 \). Type: D.

Step3: Identify Reaction Type (3)

Reaction: \( \text{NaF} + \text{Br}_2 ightarrow \text{NaBr} + \text{F}_2 \). Single Replacement (SR): Br replaces F in NaF. Balance: Left F=1, Br=2; Right F=2, Br=1. Multiply NaF by 2: \( 2\text{NaF} \) (F=2, Na=2). Multiply NaBr by 2: \( 2\text{NaBr} \) (Br=2, Na=2). Now: \( 2\text{NaF} + \text{Br}_2 ightarrow 2\text{NaBr} + \text{F}_2 \). Type: SR.

Step4: Identify Reaction Type (4)

Reaction: \( \text{SO}_2 + \text{Li}_2\text{Se} ightarrow \text{SSe}_2 + \text{Li}_2\text{O} \). Double Replacement (DR): Ions swap. Balance: Check atoms. S:1, O:2, Li:2, Se:1 (left); S:1, Se:2, Li:2, O:1 (right). Multiply \( \text{SO}_2 \) by 1, \( \text{Li}_2\text{Se} \) by 2: \( \text{SO}_2 + 2\text{Li}_2\text{Se} ightarrow \text{SSe}_2 + 2\text{Li}_2\text{O} \). Now O:2 (left), O:2 (right); Se:2 (left), Se:2 (right). Type: DR.

Step5: Identify Reaction Type (5)

Reaction: \( \text{PbSO}_4 ightarrow \text{PbSO}_3 + \text{O}_2 \). Decomposition (D): \( \text{PbSO}_4 \) breaks into \( \text{PbSO}_3 \) and \( \text{O}_2 \). Balance: Left O=4; Right O=3 (in \( \text{PbSO}_3 \)) + 2 (in \( \text{O}_2 \)) = 5? Wait, correct: \( \text{PbSO}_4 \) has 4 O, \( \text{PbSO}_3 \) has 3 O, so \( \text{O}_2 \) needs 1 O? No, balance O: Multiply \( \text{PbSO}_4 \) by 2, \( \text{PbSO}_3 \) by 2, \( \text{O}_2 \) by 1: \( 2\text{PbSO}_4 ightarrow 2\text{PbSO}_3 + \text{O}_2 \). Now O: 8 (left) = 6 (2×3) + 2 (1×2). Type: D.

Step6: Identify Reaction Type (6)

Reaction: \( \text{Al} + \text{Fe}_2\text{O}_3 ightarrow \text{Fe} + \text{Al}_2\text{O}_3 \). Single Replacement (SR): Al replaces Fe. Balance: Use coefficients. \( \text{Fe}_2\text{O}_3 \) has 2 Fe, \( \text{Al}_2\text{O}_3 \) has 2 Al. So: \( 2\text{Al} + \text{Fe}_2\text{O}_3 ightarrow 2\text{Fe} + \text{Al}_2\text{O}_3 \). Type: SR.

Step7: Identify Reaction Type (7)

Reaction: \( \text{SF}_6 ightarrow \text{S} + \text{F}_2 \). Decomposition (D): \( \text{SF}_6 \) breaks into S and \( \text{F}_2 \). Balance: Left F=6; Right F=2. Multiply \( \text{F}_2 \) by 3: \( \text{SF}_6 ightarrow \text{S} + 3\text{F}_2 \). Type: D.

Step8: Identify Reaction Type (8)

Reaction: \( \text{C}_2\text{H}_4 + \text{O}_2 ightarrow \text{CO}_2 + \text{H}_2\text{O} \). Combustion (C) of a hydrocarbon. Balance: \( \text{C}_2\text{H}_4 \) has 2 C, 4 H. Multiply \( \text{CO}_2 \) by 2 (C=2), \( \text{H}_2\text{O} \) by 2 (H=4). Now O: 2×2 + 2×1 = 6; multiply \( \text{O}_2 \) by 3: \( \text{C}_2\text{H}_4 + 3\text{O}_2 ightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} \). Type: C.

Step9: Identify Reaction Type (9)

Reaction: \( \text{KClO}_3 ightarrow \text{KCl} + \text{O}_2 \). Decomposition (D): \( \text{KClO}_3 \) breaks into KCl and \( \text{O}_2 \). Balance: Multiply \( \text{KClO}_3 \) by 2, \( \text{KCl} \) by 2, \( \text{O}_2 \) by 3: \( 2\text{KClO}_3 ightarrow 2\text{KCl} + 3\text{O}_2 \). Type: D.

Step10: Identify Reaction Type (10)

Reaction: \( \text{Pb(NO}_3\text{)}_2 + \text{NaCl} ightarrow \text{NaNO}_3 + \text{PbCl}_2 \). Double Replacement (DR): Ions swap. Balance: \( \text{PbCl}_2 \) has 2 Cl, so multiply NaCl by 2: \( 2\text{NaCl} \). Then multiply \( \text{NaNO}_3 \) by 2: \( \text{Pb(NO}_3\text{)}_2 + 2\text{NaCl} ightarrow 2\text{NaNO}_3 + \text{PbCl}_2 \). Type: DR.

Step11: Reaction Type (11)

The diagram shows C (black) and \( \text{O}_2 \) (white) forming \( \text{CO}_2 \)? Wait, the left has C (black - black circle? Wait, key: white=O, black=Carbon. So left: C (black - single? Wait, the first molecule: black and white? Wait, the diagram: first is black (C) and white (O) as \( \text{CO} \)? Wait, no: first reactant: black (C) and white (O) – maybe \( \text{CO} \)? Second reactant: \( \text{O}_2 \) (two white circles). Product: \( \text{CO}_2 \) (black, two white). Wait, no: the reaction is \( \text{CO} + \text{O}_2 ightarrow \text{CO}_2 \)? Wait, no, the first reactant: black (C) and white (O) – maybe \( \text{C} \) (black) and \( \text{O}_2 \) (white)? Wait, the first reactant: black (C) and white (O) – a molecule like \( \text{CO} \)? No, the key: white=Oxygen, black=Carbon. So first reactant: black (C) and white (O) – \( \text{CO} \)? Second reactant: \( \text{O}_2 \) (two white circles). Product: \( \text{CO}_2 \) (black, two white). Wait, the reaction is synthesis: combining \( \text{CO} \) and \( \text{O}_2 \) to \( \text{CO}_2 \)? No, maybe the first reactant is \( \text{C} \) (black) and \( \text{O}_2 \) (white). Wait, the diagram: left side: \( \text{CO} \) (black and white) + \( \text{O}_2 \) (two white) → \( \text{CO}_2 \) (black and two white). Wait, no, the first reactant: black (C) and white (O) – \( \text{CO} \), second: \( \text{O}_2 \), product: \( \text{CO}_2 \). But the type: synthesis, as two substances form one. Wait, no, the reaction shown is \( \text{CO} + \text{O}_2 ightarrow \text{CO}_2 \)? No, maybe the first reactant is \( \text{C} \) (black) and \( \text{O}_2 \) (white). Wait, the key: white=Oxygen, black=Carbon. So first reactant: black (C) and white (O) – a molecule with C and O (like \( \text{CO} \)), second: \( \text{O}_2 \) (two O), product: \( \text{CO}_2 \) (C and two O). So the reaction is synthesis (combining \( \text{CO} \) and \( \text{O}_2 \) to \( \text{CO}_2 \))? Wait, no, maybe the first reactant is \( \text{C} \) (black) and \( \text{O}_2 \) (white). Wait, the diagram: first reactant: black (C) and white (O) – \( \text{CO} \), second: \( \text{O}_2 \), product: \( \text{CO}_2 \). So type: synthesis.

Step12: Chemical Formula (12)

From the diagram: First reactant: \( \text{CO} \) (black=Carbon, white=Oxygen: one C, one O). Second reactant: \( \text{O}_2 \) (two O). Product: \( \text{CO}_2 \) (one C, two O). So formula: \( \text{CO} + \text{O}_2 ightarrow \text{CO}_2 \).

Step13: Balance Equation (13)

Equation: \( \text{CO} + \text{O}_2 ightarrow \text{CO}_2 \). Balance O: Left O=1 + 2 = 3; Right O=2. Multiply \( \text{CO} \) by 2: \( 2\text{CO} + \text{O}_2 ightarrow 2\text{CO}_2 \). Now O: 2 + 2 = 4; Right O=4. Balanced.

Answer (for each part, e.g., 1):

1) \( 2\text{K} + \text{Cl}_2 ightarrow 2\text{KCl} \), Type: Synthesis (S)
2) \( 2\text{NH}_3 ightarrow \text{N}_2 + 3\text{H}_2 \), Type: Decomposition (D)
3) \( 2\text{NaF} + \text{Br}_2 ightarrow 2\text{NaBr} + \text{F}_2 \), Type: Single Replacement (SR)
4) \( \text{SO}_2 + 2\text{Li}_2\text{Se} ightarrow \text{SSe}_2 + 2\text{Li}_2\text{O} \), Type: Double Replacement (DR)
5) \( 2\text{PbSO}_4 ightarrow 2\text{PbSO}_3 + \text{O}_2 \), Type: Decomposition (D)
6) \( 2\text{Al} + \text{Fe}_2\text{O}_3 ightarrow 2\text{Fe} + \text{Al}_2\text{O}_3 \), Type: Single Replacement (SR)
7) \( \text{SF}_6 ightarrow \text{S} + 3\text{F}_2 \), Type: Decomposition (D)
8) \( \text{C}_2\text{H}_4 + 3\text{O}_2 ightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} \), Type: Combustion (C)
9) \( 2\text{KClO}_3 ightarrow 2\text{KCl} + 3\text{O}_2 \), Type: Decomposition (D)
10) \( \text{Pb(NO}_3\text{)}_2 + 2\text{NaCl} ightarrow 2\text{NaNO}_3 + \text{PbCl}_2 \), Type: Double Replacement (DR)
11) Type: Synthesis (S) (combining reactants to form one product)
12) \( \text{CO} + \text{O}_2 ightarrow \text{CO}_2 \)
13) \( 2\text{CO} + \text{O}_2 ightarrow 2\text{CO}_2 \)

(Note: For each sub-question, the answer includes the balanced equation and reaction type as applicable.)

Answer:

(for each part, e.g., 1):

1. \( 2\text{K} + \text{Cl}_2

ightarrow 2\text{KCl} \), Type: Synthesis (S)

1. \( 2\text{NH}_3

ightarrow \text{N}_2 + 3\text{H}_2 \), Type: Decomposition (D)

1. \( 2\text{NaF} + \text{Br}_2

ightarrow 2\text{NaBr} + \text{F}_2 \), Type: Single Replacement (SR)

1. \( \text{SO}_2 + 2\text{Li}_2\text{Se}

ightarrow \text{SSe}_2 + 2\text{Li}_2\text{O} \), Type: Double Replacement (DR)

1. \( 2\text{PbSO}_4

ightarrow 2\text{PbSO}_3 + \text{O}_2 \), Type: Decomposition (D)

1. \( 2\text{Al} + \text{Fe}_2\text{O}_3

ightarrow 2\text{Fe} + \text{Al}_2\text{O}_3 \), Type: Single Replacement (SR)

1. \( \text{SF}_6

ightarrow \text{S} + 3\text{F}_2 \), Type: Decomposition (D)

1. \( \text{C}_2\text{H}_4 + 3\text{O}_2

ightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} \), Type: Combustion (C)

1. \( 2\text{KClO}_3

ightarrow 2\text{KCl} + 3\text{O}_2 \), Type: Decomposition (D)

1. \( \text{Pb(NO}_3\text{)}_2 + 2\text{NaCl}

ightarrow 2\text{NaNO}_3 + \text{PbCl}_2 \), Type: Double Replacement (DR)

1. Type: Synthesis (S) (combining reactants to form one product)
2. \( \text{CO} + \text{O}_2

ightarrow \text{CO}_2 \)

1. \( 2\text{CO} + \text{O}_2

ightarrow 2\text{CO}_2 \)

(Note: For each sub-question, the answer includes the balanced equation and reaction type as applicable.)