Question

for the chemical equation below, list the stoichiometric coefficients in the order they would appear in the equation (1s included)
burning of
_c₈h₁₈ + _o₂ ==> _co₂ + _h₂o

○ 2 : 25 : 16 : 18

○ 1 : 8 : 8 : 9

○ 2 : 25 : 8 : 9

○ 2 : 50 : 16 : 18

○ 1 : 12 : 8 : 9

Explanation:

Step1: Balance Carbon Atoms

In \( \text{C}_8\text{H}_{18} \), there are 8 C atoms. So, \( \text{CO}_2 \) should have a coefficient of 8 (since each \( \text{CO}_2 \) has 1 C). Wait, no, let's start properly. The unbalanced equation is \( \text{C}_8\text{H}_{18} + \text{O}_2
ightarrow \text{CO}_2 + \text{H}_2\text{O} \).

First, balance C: \( \text{C}_8\text{H}_{18} \) has 8 C, so \( \text{CO}_2 \) gets 8? Wait, no, let's use the standard method. Let's let the coefficient of \( \text{C}_8\text{H}_{18} \) be 2 to make H even. So \( 2\text{C}_8\text{H}_{18} \) has 16 C and 36 H.

Step2: Balance Carbon

16 C means \( \text{CO}_2 \) has coefficient 16 (since each \( \text{CO}_2 \) has 1 C). Wait, no, \( 2\text{C}_8\text{H}_{18} \) has 16 C, so \( \text{CO}_2 \) coefficient is 16? Wait, no, \( \text{C}_8\text{H}_{18} \) has 8 C per molecule. So if we take 2 molecules, 16 C. So \( \text{CO}_2 \) needs 16. Then H: 2 \( \text{C}_8\text{H}_{18} \) has 36 H, so \( \text{H}_2\text{O} \) has coefficient 18 (since each \( \text{H}_2\text{O} \) has 2 H, 182=36). Now O: on the right, \( 16\text{CO}_2 \) has 32 O, \( 18\text{H}_2\text{O} \) has 18 O, total 50 O. So \( \text{O}_2 \) needs 25 (since each \( \text{O}_2 \) has 2 O, 252=50). So the balanced equation is \( 2\text{C}_8\text{H}_{18} + 25\text{O}_2
ightarrow 16\text{CO}_2 + 18\text{H}_2\text{O} \). So coefficients are 2,25,16,18.

Wait, but let's check the options. The first option is 2:25:16:18. Let's verify:

Left: C: 28=16; H: 218=36; O: 25*2=50.

Right: C:161=16; H:182=36; O:162 + 181=32+18=50. Balanced. So the correct coefficients are 2,25,16,18.

Answer:

A. 2 : 25 : 16 : 18