Question

acetaldehyde is a substance commonly used in chemical factories. it is often produced by combining ethylene gas and oxygen gas. consider the following reaction: 4 molecules of ethylene (c₂h₄) react with 2 molecules of oxygen (o₂) to produce some amount of acetaldehyde (c₂h₄o). complete the table below. table: chemical element (h, c, o) | number of atoms in the reaction (blanks for each element) during this reaction, how many molecules of acetaldehyde (c₂h₄o) are produced?

Explanation:

Part 1: Completing the table (number of atoms)

For Hydrogen (H):

Step1: Calculate H from $C_2H_4$

Each $C_2H_4$ has 4 H atoms. There are 4 molecules of $C_2H_4$. So total H from $C_2H_4$: $4\times4 = 16$.
<Expression> $4 \text{ molecules} \times 4 \text{ H atoms/molecule} = 16$</Expression>

Step2: Check H in product (no H from $O_2$)

Since $O_2$ has no H, total H atoms in reaction: 16.
<Expression> Total H = 16</Expression>

For Carbon (C):

Step1: Calculate C from $C_2H_4$

Each $C_2H_4$ has 2 C atoms. 4 molecules of $C_2H_4$: $4\times2 = 8$.
<Expression> $4 \text{ molecules} \times 2 \text{ C atoms/molecule} = 8$</Expression>

Step2: No C from $O_2$

Total C atoms: 8.
<Expression> Total C = 8</Expression>

For Oxygen (O):

Step1: Calculate O from $O_2$

Each $O_2$ has 2 O atoms. 2 molecules of $O_2$: $2\times2 = 4$.
<Expression> $2 \text{ molecules} \times 2 \text{ O atoms/molecule} = 4$</Expression>

Step2: No O from $C_2H_4$ (before reaction)

Total O atoms: 4.
<Expression> Total O = 4</Expression>

Part 2: Molecules of acetaldehyde ($C_2H_4O$) produced

Step1: Use conservation of atoms (or balanced reaction)

Let the number of $C_2H_4O$ molecules be $x$.

• For C: Each $C_2H_4O$ has 2 C. Total C from product: $2x$. We know total C is 8 (from reactants). So $2x = 8 \implies x = 4$.
• Verify with H: Each $C_2H_4O$ has 4 H. $4x = 4\times4 = 16$ (matches total H).
• Verify with O: Each $C_2H_4O$ has 1 O. $x = 4$ gives 4 O (matches total O).

<Expression> $2x = 8 \implies x = \frac{8}{2} = 4$</Expression>

Table Completion Answers:

• H: 16
• C: 8
• O: 4

Molecules of Acetaldehyde:

4

Answer:

Part 1: Completing the table (number of atoms)

For Hydrogen (H):

Step1: Calculate H from $C_2H_4$

Each $C_2H_4$ has 4 H atoms. There are 4 molecules of $C_2H_4$. So total H from $C_2H_4$: $4\times4 = 16$.
<Expression> $4 \text{ molecules} \times 4 \text{ H atoms/molecule} = 16$</Expression>

Step2: Check H in product (no H from $O_2$)

Since $O_2$ has no H, total H atoms in reaction: 16.
<Expression> Total H = 16</Expression>

For Carbon (C):

Step1: Calculate C from $C_2H_4$

Each $C_2H_4$ has 2 C atoms. 4 molecules of $C_2H_4$: $4\times2 = 8$.
<Expression> $4 \text{ molecules} \times 2 \text{ C atoms/molecule} = 8$</Expression>

Step2: No C from $O_2$

Total C atoms: 8.
<Expression> Total C = 8</Expression>

For Oxygen (O):

Step1: Calculate O from $O_2$

Each $O_2$ has 2 O atoms. 2 molecules of $O_2$: $2\times2 = 4$.
<Expression> $2 \text{ molecules} \times 2 \text{ O atoms/molecule} = 4$</Expression>

Step2: No O from $C_2H_4$ (before reaction)

Total O atoms: 4.
<Expression> Total O = 4</Expression>

Part 2: Molecules of acetaldehyde ($C_2H_4O$) produced

Step1: Use conservation of atoms (or balanced reaction)

Let the number of $C_2H_4O$ molecules be $x$.

• For C: Each $C_2H_4O$ has 2 C. Total C from product: $2x$. We know total C is 8 (from reactants). So $2x = 8 \implies x = 4$.
• Verify with H: Each $C_2H_4O$ has 4 H. $4x = 4\times4 = 16$ (matches total H).
• Verify with O: Each $C_2H_4O$ has 1 O. $x = 4$ gives 4 O (matches total O).

<Expression> $2x = 8 \implies x = \frac{8}{2} = 4$</Expression>

Table Completion Answers:

• H: 16
• C: 8
• O: 4

Molecules of Acetaldehyde:

4