To balance the given reaction [tex]\( \text{Fe}_2\text{O}_3 + \text{H}_2 \rightarrow \text{Fe} + \text{H}_2\text{O} \)[/tex], we need to ensure that the number of each type of atom on the reactant side is equal to the number on the product side. Let's go through this step-wise.
1. Write the unbalanced equation:
[tex]\[
\text{Fe}_2\text{O}_3 + \text{H}_2 \rightarrow \text{Fe} + \text{H}_2\text{O}
\][/tex]
2. Count the number of atoms for each element on both sides:
- On the reactant side:
- Fe: 2
- O: 3
- H: 2
- On the product side:
- Fe: 1
- O: 1
- H: 2 (in [tex]\( \text{H}_2\text{O} \)[/tex])
3. Balance the iron (Fe) atoms first:
Since there are 2 Fe atoms on the reactant side in [tex]\( \text{Fe}_2\text{O}_3 \)[/tex], we need 2 Fe atoms on the product side.
[tex]\[
\text{Fe}_2\text{O}_3 + \text{H}_2 \rightarrow 2\text{Fe} + \text{H}_2\text{O}
\][/tex]
4. Balance the oxygen (O) atoms:
There are 3 oxygen atoms in [tex]\( \text{Fe}_2\text{O}_3 \)[/tex] on the reactant side, so we need 3 [tex]\( \text{H}_2\text{O} \)[/tex] molecules to get 3 oxygen atoms on the product side.
[tex]\[
\text{Fe}_2\text{O}_3 + \text{H}_2 \rightarrow 2\text{Fe} + 3\text{H}_2\text{O}
\][/tex]
5. Balance the hydrogen (H) atoms:
On the product side, we have 3 molecules of [tex]\( \text{H}_2\text{O} \)[/tex], each contributing 2 hydrogen atoms, making a total of 6 hydrogen atoms. Therefore, we need 3 molecules of [tex]\( \text{H}_2 \)[/tex] on the reactant side to balance the hydrogen atoms.
[tex]\[
\text{Fe}_2\text{O}_3 + 3\text{H}_2 \rightarrow 2\text{Fe} + 3\text{H}_2\text{O}
\][/tex]
Now, the final balanced equation is:
[tex]\[
\text{Fe}_2\text{O}_3 + 3\text{H}_2 \rightarrow 2\text{Fe} + 3\text{H}_2\text{O}
\][/tex]
Thus, the coefficients [tex]\( a, b, c, \)[/tex] and [tex]\( d \)[/tex] respectively are [tex]\(1, 3, 2, 3\)[/tex].
The correct answer is:
(C) [tex]\( 1, 3, 2, 3 \)[/tex]
(D) [tex]\( 1, 3, 2, 3\)[/tex]
