Question

1. when the equation 2.fe₂o₃ + _ h₂ → _ fe + _ h₂o is balanced using whole - number coefficients. he has a coefficient of

a. 1
b. 6
c. 3
d. 2

Explanation:

Step1: Analyze Fe atoms

Left side: \(2\ce{Fe2O3}\) has \(2\times2 = 4\) Fe atoms. So Fe on right (product) should have coefficient 4? Wait, no, wait the original equation: Wait, the given equation is \(2\ce{Fe2O3}+\_\ce{H2}
ightarrow\_\ce{Fe}+\_\ce{H2O}\). Wait, no, maybe a typo? Wait, no, let's re - examine. Wait, \( \ce{Fe2O3}\) has 2 Fe per formula unit. If left has \(2\ce{Fe2O3}\), that's \(2\times2 = 4\) Fe. So Fe on right should be 4? But the options are 1, 6, 3, 2. Wait, maybe the original equation is \( \ce{Fe2O3}+\_\ce{H2}
ightarrow\_\ce{Fe}+\_\ce{H2O}\) (without the 2 in front of \( \ce{Fe2O3}\) initially). Wait, maybe the user made a typo, but let's assume the correct unbalanced equation is \( \ce{Fe2O3}+\ce{H2}
ightarrow\ce{Fe}+\ce{H2O}\).

Step2: Balance Fe

Left: 2 Fe (in \( \ce{Fe2O3}\)), right: 1 Fe. So Fe coefficient on right should be 2? No, wait, let's do proper balancing.

1. Balance Fe: \( \ce{Fe2O3}\) has 2 Fe, so put 2 in front of Fe: \( \ce{Fe2O3}+\ce{H2}

ightarrow2\ce{Fe}+\ce{H2O}\)

1. Balance O: \( \ce{Fe2O3}\) has 3 O, so put 3 in front of \( \ce{H2O}\): \( \ce{Fe2O3}+\ce{H2}

ightarrow2\ce{Fe}+3\ce{H2O}\)

1. Balance H: \(3\ce{H2O}\) has 6 H, so put 3 in front of \( \ce{H2}\): \( \ce{Fe2O3}+3\ce{H2}

ightarrow2\ce{Fe}+3\ce{H2O}\)

But the given equation in the problem has \(2\ce{Fe2O3}\) on the left. Let's adjust for that. If left is \(2\ce{Fe2O3}\), then:

• Fe: \(2\times2 = 4\), so Fe on right: 4
• O: \(2\times3 = 6\), so \( \ce{H2O}\) coefficient: 6 (since each \( \ce{H2O}\) has 1 O)
• H: \(6\ce{H2O}\) has 12 H, so \( \ce{H2}\) coefficient: 6 (since each \( \ce{H2}\) has 2 H, \(12\div2 = 6\))
• Fe: 4 (from \(2\ce{Fe2O3}\), \(2\times2 = 4\))

Wait, but the question is about the coefficient of \( \ce{H2}\)? No, the question says "He has a coefficient of" (maybe a typo, maybe "H₂ has a coefficient of"). Wait, the options are A.1, B.6, C.3, D.2.

Wait, let's go back to the original problem as presented: "When the equation \(2\ce{Fe2O3}+\_\ce{H2}
ightarrow\_\ce{Fe}+\_\ce{H2O}\) is balanced using whole - number coefficients. H₂ has a coefficient of"

1. Balance Fe: \(2\ce{Fe2O3}\) has \(2\times2 = 4\) Fe, so Fe on right: 4
2. Balance O: \(2\ce{Fe2O3}\) has \(2\times3 = 6\) O, so \( \ce{H2O}\) coefficient: 6 (since each \( \ce{H2O}\) has 1 O)
3. Balance H: \(6\ce{H2O}\) has \(6\times2 = 12\) H, so \( \ce{H2}\) coefficient: \(12\div2 = 6\)

Answer:

B. 6