Balance the chemical equation for the reaction between ethylene and oxygen to form acetaldehyde, given 2 molecules of ethylene and 1 molecule of oxygen react.

The reactants are 2 molecules of ethylene ($\text{C}_2\text{H}_4$) and 1 molecule of oxygen ($\text{O}_2$). The product is acetaldehyde ($\text{C}_2\text{H}_4\text{O}$).

• For Carbon (C): In $2\text{C}_2\text{H}_4$, there are $2 \times 2 = 4$ carbon atoms.
• For Hydrogen (H): In $2\text{C}_2\text{H}_4$, there are $2 \times 4 = 8$ hydrogen atoms.
• For Oxygen (O): In $\text{O}_2$, there are $1 \times 2 = 2$ oxygen atoms.

According to the law of conservation of atoms, the number of atoms of each element must be the same on both sides of the equation. Let $x$ be the stoichiometric coefficient for acetaldehyde.

• For Carbon (C): Reactants have 4 C atoms. In $x\text{C}_2\text{H}_4\text{O}$, there are $x \times 2$ carbon atoms. So, $2x = 4 \Rightarrow x = 2$.
• For Hydrogen (H): Reactants have 8 H atoms. In $x\text{C}_2\text{H}_4\text{O}$, there are $x \times 4$ hydrogen atoms. So, $4x = 8 \Rightarrow x = 2$.
• For Oxygen (O): Reactants have 2 O atoms. In $x\text{C}_2\text{H}_4\text{O}$, there are $x \times 1$ oxygen atoms. So, $1x = 2 \Rightarrow x = 2$.

Since $x=2$ for all elements, the balanced equation is:

Now, count the atoms in the products with $x=2$:

• For Carbon (C): In $2\text{C}_2\text{H}_4\text{O}$, there are $2 \times 2 = 4$ carbon atoms.
• For Hydrogen (H): In $2\text{C}_2\text{H}_4\text{O}$, there are $2 \times 4 = 8$ hydrogen atoms.
• For Oxygen (O): In $2\text{C}_2\text{H}_4\text{O}$, there are $2 \times 1 = 2$ oxygen atoms.