Question

when you balance the equation c₅h₁₂(l) + o₂(g) → co₂(g) + h₂o(g), the correct coefficient of o₂ will be: 1) 5 2) 6 3) 11 4) 8

Explanation:

Step1: Balance Carbon Atoms

In the reactant, we have \( \ce{C5H12} \), so there are 5 carbon atoms. Thus, we need 5 \( \ce{CO2} \) molecules on the product side. So the equation becomes \( \ce{C5H12(l) + O2(g) -> 5CO2(g) + H2O(g)} \).

Step2: Balance Hydrogen Atoms

In \( \ce{C5H12} \), there are 12 hydrogen atoms. Each \( \ce{H2O} \) has 2 hydrogen atoms, so we need \( \frac{12}{2} = 6 \) \( \ce{H2O} \) molecules. Now the equation is \( \ce{C5H12(l) + O2(g) -> 5CO2(g) + 6H2O(g)} \).

Step3: Balance Oxygen Atoms

On the product side, we have \( 5 \times 2 + 6 \times 1 = 10 + 6 = 16 \) oxygen atoms from \( \ce{CO2} \) and \( \ce{H2O} \). Since each \( \ce{O2} \) molecule has 2 oxygen atoms, the number of \( \ce{O2} \) molecules needed is \( \frac{16}{2} = 8 \)? Wait, no, wait. Wait, let's recalculate. Wait, \( 5\ce{CO2} \) has \( 5 \times 2 = 10 \) O, \( 6\ce{H2O} \) has \( 6 \times 1 = 6 \) O, total O on product side is \( 10 + 6 = 16 \). So \( \ce{O2} \) has 2 O per molecule, so number of \( \ce{O2} \) is \( 16 / 2 = 8 \)? Wait, no, that's not right. Wait, maybe I made a mistake. Wait, let's do it again. Wait, the formula is \( \ce{C5H12 + O2 -> CO2 + H2O} \).

Wait, carbon: 5 on left, so 5 \( \ce{CO2} \). Hydrogen: 12 on left, so 6 \( \ce{H2O} \) (since 62=12). Now oxygen: on right, \( 5*2 + 6*1 = 10 + 6 = 16 \). So \( \ce{O2} \) has 2 O, so number of \( \ce{O2} \) is \( 16/2 = 8 \)? But wait, the options have 8 as option 4, but wait, maybe I messed up. Wait, no, wait, the correct balancing for \( \ce{C5H12} \) (pentane) combustion is \( \ce{C5H12 + 8O2 -> 5CO2 + 6H2O} \)? Wait, no, wait, let's check again. Wait, \( 5\ce{CO2} \) is 5 C, 10 O. \( 6\ce{H2O} \) is 12 H, 6 O. Total O on right: 16. So \( \ce{O2} \) is 8, because 82=16. So the coefficient of \( \ce{O2} \) is 8? But wait, the options are 5,6,11,8. Wait, maybe I made a mistake. Wait, no, wait, pentane combustion: the correct balanced equation is \( \ce{C5H12 + 8O2 -> 5CO2 + 6H2O} \)? Wait, no, wait, let's calculate again. Wait, 5 C, 12 H. So \( 5\ce{CO2} \) (5 C), \( 6\ce{H2O} \) (12 H). Then O: 52 + 61 = 10 + 6 = 16. So \( \ce{O2} \) is 16/2 = 8. So the coefficient of \( \ce{O2} \) is 8? But wait, the options have 8 as option 4. But wait, maybe I made a mistake. Wait, no, let's check the standard combustion reaction for pentane: \( \ce{C5H12 + 8O2 -> 5CO2 + 6H2O} \). Yes, that's correct. So the coefficient of \( \ce{O2} \) is 8. So the correct option is 4) 8. Wait, but wait, the initial calculation I thought 16/2=8, which is correct. So the answer is 8, option 4.

Wait, but wait, maybe I messed up. Wait, let's check again. Carbon: 5 on left, 5 on right (5 CO2). Hydrogen: 12 on left, 12 on right (6 H2O, 62=12). Oxygen: on right, 52 (from CO2) + 6*1 (from H2O) = 10 + 6 = 16. So O2 has 2 O per molecule, so number of O2 is 16/2 = 8. So yes, coefficient of O2 is 8. So the correct option is 4) 8.

Wait, but the options are 1)5, 2)6, 3)11, 4)8. So the answer is 4) 8.

Answer:

1. 8