7. predict the products of each reaction. then balance each chemical equation. state the type of reaction.
$ce{al + br_{2} -}$
$ce{cu + agno_{3} -}$
$ce{na_{2}o + h_{2}o -}$
$ce{mg + hcl -}$
$ce{k + o_{2} -}$
$ce{al_{2}o_{3} -}$
$ce{ca + h_{2}o -}$
with lead (ii) acetate to produce a bright yellow precipita

Question

7. predict the products of each reaction. then balance each chemical equation. state the type of reaction.
$ce{al + br_{2} ->}$
$ce{cu + agno_{3} ->}$
$ce{na_{2}o + h_{2}o ->}$
$ce{mg + hcl ->}$
$ce{k + o_{2} ->}$
$ce{al_{2}o_{3} ->}$
$ce{ca + h_{2}o ->}$
with lead (ii) acetate to produce a bright yellow precipita

Accepted Answer

### 1. Reaction: $\boldsymbol{\ce{Al + Br_{2}
ightarrow}}$
# Explanation:
## Step 1: Predict Products (Synthesis Reaction)
Aluminum (Al) reacts with bromine ($\ce{Br_{2}}$) to form aluminum bromide ($\ce{AlBr_{3}}$) (synthesis: metal + non - metal $
ightarrow$ ionic compound).
Unbalanced: $\ce{Al + Br_{2}
ightarrow AlBr_{3}}$
## Step 2: Balance the Equation
- Balance Br: There are 2 Br on left, 3 on right. LCM of 2 and 3 is 6. So put 3 in front of $\ce{Br_{2}}$ and 2 in front of $\ce{AlBr_{3}}$.
  $\ce{Al + 3Br_{2}
ightarrow 2AlBr_{3}}$
- Balance Al: Now there are 2 Al on right, so put 2 in front of Al.
  $\ce{2Al + 3Br_{2}
ightarrow 2AlBr_{3}}$
## Step 3: Reaction Type
Synthesis (combination) reaction (two substances combine to form one).

### 2. Reaction: $\boldsymbol{\ce{Cu + AgNO_{3}
ightarrow}}$
# Explanation:
## Step 1: Predict Products (Single - Displacement)
Copper (Cu) is more reactive than silver (Ag), so Cu displaces Ag from $\ce{AgNO_{3}}$. Products are $\ce{Cu(NO_{3})_{2}}$ and Ag.
Unbalanced: $\ce{Cu + AgNO_{3}
ightarrow Cu(NO_{3})_{2} + Ag}$
## Step 2: Balance the Equation
- Balance $\ce{NO_{3}^{-}}$: 2 on right, so put 2 in front of $\ce{AgNO_{3}}$.
  $\ce{Cu + 2AgNO_{3}
ightarrow Cu(NO_{3})_{2} + Ag}$
- Balance Ag: 2 on left, so put 2 in front of Ag.
  $\ce{Cu + 2AgNO_{3}
ightarrow Cu(NO_{3})_{2} + 2Ag}$
## Step 3: Reaction Type
Single - displacement reaction (one element displaces another in a compound).

### 3. Reaction: $\boldsymbol{\ce{Na_{2}O + H_{2}O
ightarrow}}$
# Explanation:
## Step 1: Predict Products (Synthesis)
Sodium oxide ($\ce{Na_{2}O}$) reacts with water ($\ce{H_{2}O}$) to form sodium hydroxide ($\ce{NaOH}$) (synthesis of a base from metal oxide and water).
Unbalanced: $\ce{Na_{2}O + H_{2}O
ightarrow NaOH}$
## Step 2: Balance the Equation
- Balance Na: 2 on left, so put 2 in front of $\ce{NaOH}$.
  $\ce{Na_{2}O + H_{2}O
ightarrow 2NaOH}$
- Check O and H: O: 2 (left: 1 in $\ce{Na_{2}O}$ + 1 in $\ce{H_{2}O}$ = 2; right: 2 in 2$\ce{NaOH}$). H: 2 (left: 2 in $\ce{H_{2}O}$; right: 2 in 2$\ce{NaOH}$). Balanced.
## Step 3: Reaction Type
Synthesis (combination) reaction (metal oxide + water $
ightarrow$ base).

### 4. Reaction: $\boldsymbol{\ce{Mg + HCl
ightarrow}}$
# Explanation:
## Step 1: Predict Products (Single - Displacement)
Magnesium (Mg) is more reactive than H, so Mg displaces H from HCl. Products are $\ce{MgCl_{2}}$ and $\ce{H_{2}}$ (gas).
Unbalanced: $\ce{Mg + HCl
ightarrow MgCl_{2} + H_{2}}$
## Step 2: Balance the Equation
- Balance Cl: 2 on right, so put 2 in front of HCl.
  $\ce{Mg + 2HCl
ightarrow MgCl_{2} + H_{2}}$
- Check Mg and H: Mg is balanced (1 on each side), H is balanced (2 on each side).
## Step 3: Reaction Type
Single - displacement reaction (metal displaces H from acid).

### 5. Reaction: $\boldsymbol{\ce{K + O_{2}
ightarrow}}$
# Explanation:
## Step 1: Predict Products (Synthesis)
Potassium (K) reacts with oxygen ($\ce{O_{2}}$) to form potassium oxide (or peroxide, but primary is $\ce{K_{2}O}$ for simple synthesis; if considering peroxide, $\ce{K_{2}O_{2}}$, but let's do $\ce{K_{2}O}$ first).
Unbalanced: $\ce{K + O_{2}
ightarrow K_{2}O}$
## Step 2: Balance the Equation
- Balance O: 2 on left, 1 on right. Put 2 in front of $\ce{K_{2}O}$.
  $\ce{K + O_{2}
ightarrow 2K_{2}O}$
- Balance K: 4 on right, so put 4 in front of K.
  $\ce{4K + O_{2}
ightarrow 2K_{2}O}$
- If we consider potassium peroxide ($\ce{K_{2}O_{2}}$):
  Unbalanced: $\ce{K + O_{2}
ightarrow K_{2}O_{2}}$
  Balance K: 2 in front of K. $\ce{2K + O_{2}
ightarrow K_{2}O_{2}}$ (this is also a valid synthesis, and is more likely as K is very reactive and forms peroxides easily. Let's use this as it's more accurate for K's reaction with $\ce{O_{2}}$ under normal conditions for simple balancing).
## Step 3: Reaction Type
Synthesis (combination) reaction.

### 6. Reaction: $\boldsymbol{\ce{Al_{2}O_{3}
ightarrow}}$
# Explanation:
## Step 1: Predict Products (Decomposition)
Aluminum oxide ($\ce{Al_{2}O_{3}}$) decomposes into aluminum (Al) and oxygen ($\ce{O_{2}}$) (decomposition of metal oxide into metal and oxygen, requires high temperature).
Unbalanced: $\ce{Al_{2}O_{3}
ightarrow Al + O_{2}}$
## Step 2: Balance the Equation
- Balance O: 3 on left, 2 on right. LCM of 3 and 2 is 6. Put 2 in front of $\ce{Al_{2}O_{3}}$ and 3 in front of $\ce{O_{2}}$.
  $\ce{2Al_{2}O_{3}
ightarrow Al + 3O_{2}}$
- Balance Al: 4 on left, so put 4 in front of Al.
  $\ce{2Al_{2}O_{3}
ightarrow 4Al + 3O_{2}}$
## Step 3: Reaction Type
Decomposition reaction (one substance breaks into two or more).

### 7. Reaction: $\boldsymbol{\ce{Ca + H_{2}O
ightarrow}}$
# Explanation:
## Step 1: Predict Products (Single - Displacement)
Calcium (Ca) is reactive with water. It displaces H from $\ce{H_{2}O}$. Products are $\ce{Ca(OH)_{2}}$ (calcium hydroxide) and $\ce{H_{2}}$ (gas).
Unbalanced: $\ce{Ca + H_{2}O
ightarrow Ca(OH)_{2} + H_{2}}$
## Step 2: Balance the Equation
- Balance O and H in $\ce{OH^{-}}$: There are 2 $\ce{OH^{-}}$ on right, so put 2 in front of $\ce{H_{2}O}$.
  $\ce{Ca + 2H_{2}O
ightarrow Ca(OH)_{2} + H_{2}}$
- Check Ca and H: Ca is balanced (1 on each side). H: 4 on left (2*2), 2 (in $\ce{Ca(OH)_{2}}$) + 2 (in $\ce{H_{2}}$) = 4. Balanced.
## Step 3: Reaction Type
Single - displacement reaction (metal displaces H from water).

# Answers (with balanced equations and reaction types):
1. $\ce{2Al + 3Br_{2}
ightarrow 2AlBr_{3}}$; Type: Synthesis
2. $\ce{Cu + 2AgNO_{3}
ightarrow Cu(NO_{3})_{2} + 2Ag}$; Type: Single - Displacement
3. $\ce{Na_{2}O + H_{2}O
ightarrow 2NaOH}$; Type: Synthesis
4. $\ce{Mg + 2HCl
ightarrow MgCl_{2} + H_{2}\uparrow}$; Type: Single - Displacement
5. $\ce{2K + O_{2}
ightarrow K_{2}O_{2}}$ (or $\ce{4K + O_{2}
ightarrow 2K_{2}O}$); Type: Synthesis
6. $\ce{2Al_{2}O_{3}
ightarrow 4Al + 3O_{2}\uparrow}$; Type: Decomposition
7. $\ce{Ca + 2H_{2}O
ightarrow Ca(OH)_{2} + H_{2}\uparrow}$; Type: Single - Displacement

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QUESTION IMAGE

Question

Explanation:

1. Reaction: $\boldsymbol{\ce{Al + Br_{2}

Step 1: Predict Products (Synthesis Reaction)

Step 2: Balance the Equation

Step 3: Reaction Type

2. Reaction: $\boldsymbol{\ce{Cu + AgNO_{3}

Step 1: Predict Products (Single - Displacement)

Step 2: Balance the Equation

Step 3: Reaction Type

3. Reaction: $\boldsymbol{\ce{Na_{2}O + H_{2}O

Step 1: Predict Products (Synthesis)

Step 2: Balance the Equation

Step 3: Reaction Type

4. Reaction: $\boldsymbol{\ce{Mg + HCl

Answer: