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: 6 molecules of ethylene (c₂h₄) react with 3 molecules of oxygen (o₂) to produce some amount of acetaldehyde (c₂h₄o). complete the table below. chemical element | number of atoms in the reaction h | c | o | 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. 6 molecules of $C_2H_4$: $6\times4 = 24$ H atoms.

Step2: No H from $O_2$ or $C_2H_4O$ (since we count reactants, but conservation: same as products, but here reactants: $C_2H_4$ has H, $O_2$ has none. So total H atoms: 24.

For Carbon (C):

Step1: Calculate C from $C_2H_4$

Each $C_2H_4$ has 2 C atoms. 6 molecules: $6\times2 = 12$ C atoms.

Step2: No C from $O_2$, so total C atoms: 12.

For Oxygen (O):

Step1: Calculate O from $O_2$

Each $O_2$ has 2 O atoms. 3 molecules: $3\times2 = 6$ O atoms.

Step2: No O from $C_2H_4$, so total O atoms: 6.

Part 2: Molecules of acetaldehyde ($C_2H_4O$)

Step1: Use conservation of atoms. Let $x$ be molecules of $C_2H_4O$.

• For C: Each $C_2H_4O$ has 2 C. So $2x = 12$ (from C atoms in reactants).

Step2: Solve for $x$

$x=\frac{12}{2}=6$.

• Check O: Each $C_2H_4O$ has 1 O. $6\times1 = 6$, which matches O atoms from reactants (3 $O_2$: 6 O).
• Check H: Each $C_2H_4O$ has 4 H. $6\times4 = 24$, which matches H atoms from reactants (6 $C_2H_4$: 24 H).

Table Completion:

Chemical element	Number of atoms in the reaction
C	12
O	6

Molecules of acetaldehyde: 6

Final Answers:

Table: H: 24, C: 12, O: 6; Acetaldehyde molecules: 6

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. 6 molecules of $C_2H_4$: $6\times4 = 24$ H atoms.

Step2: No H from $O_2$ or $C_2H_4O$ (since we count reactants, but conservation: same as products, but here reactants: $C_2H_4$ has H, $O_2$ has none. So total H atoms: 24.

For Carbon (C):

Step1: Calculate C from $C_2H_4$

Each $C_2H_4$ has 2 C atoms. 6 molecules: $6\times2 = 12$ C atoms.

Step2: No C from $O_2$, so total C atoms: 12.

For Oxygen (O):

Step1: Calculate O from $O_2$

Each $O_2$ has 2 O atoms. 3 molecules: $3\times2 = 6$ O atoms.

Step2: No O from $C_2H_4$, so total O atoms: 6.

Part 2: Molecules of acetaldehyde ($C_2H_4O$)

Step1: Use conservation of atoms. Let $x$ be molecules of $C_2H_4O$.

• For C: Each $C_2H_4O$ has 2 C. So $2x = 12$ (from C atoms in reactants).

Step2: Solve for $x$

$x=\frac{12}{2}=6$.

• Check O: Each $C_2H_4O$ has 1 O. $6\times1 = 6$, which matches O atoms from reactants (3 $O_2$: 6 O).
• Check H: Each $C_2H_4O$ has 4 H. $6\times4 = 24$, which matches H atoms from reactants (6 $C_2H_4$: 24 H).

Table Completion:

Chemical element	Number of atoms in the reaction
C	12
O	6

Molecules of acetaldehyde: 6

Final Answers:

Table: H: 24, C: 12, O: 6; Acetaldehyde molecules: 6