To balance the chemical equation for the combustion of sugar, we need to ensure that the number of each type of atom on the reactant side is equal to the number of each type of atom on the product side. The unbalanced equation is:
[tex]\[ C_6H_{12}O_6(s) + O_2(g) \rightarrow CO_2(g) + H_2O(l) \][/tex]
Let’s determine the appropriate coefficients step-by-step:
1. Balancing Carbon (C) Atoms:
- On the reactants side, we have 6 carbon atoms in [tex]\( C_6H_{12}O_6 \)[/tex].
- Therefore, we need 6 molecules of [tex]\( CO_2 \)[/tex] on the products side to balance the carbon atoms:
[tex]\[ C_6H_{12}O_6 + O_2 \rightarrow 6CO_2 + H_2O \][/tex]
2. Balancing Hydrogen (H) Atoms:
- On the reactants side, we have 12 hydrogen atoms in [tex]\( C_6H_{12}O_6 \)[/tex].
- Each water molecule ([tex]\( H_2O \)[/tex]) contains 2 hydrogen atoms, so we need 6 molecules of [tex]\( H_2O \)[/tex] to balance the hydrogen atoms:
[tex]\[ C_6H_{12}O_6 + O_2 \rightarrow 6CO_2 + 6H_2O \][/tex]
3. Balancing Oxygen (O) Atoms:
- On the reactants side, we have:
- 6 oxygen atoms in [tex]\( C_6H_{12}O_6 \)[/tex].
- Let’s denote the oxygen atoms in [tex]\( O_2 \)[/tex] with the variable [tex]\( x \)[/tex].
- On the products side, we have:
- [tex]\( 6 \times 2 = 12 \)[/tex] oxygen atoms in [tex]\( 6CO_2 \)[/tex].
- [tex]\( 6 \times 1 = 6 \)[/tex] oxygen atoms in [tex]\( 6H_2O \)[/tex].
- In total, 18 oxygen atoms on the products side.
- Hence, the number of oxygen atoms from [tex]\( O_2 \)[/tex] should be [tex]\( 18 - 6 = 12 \)[/tex] (since 6 oxygen atoms are already present in [tex]\( C_6H_{12}O_6 \)[/tex]).
- Each [tex]\( O_2 \)[/tex] molecule provides 2 oxygen atoms, meaning we need [tex]\(\frac{12}{2} = 6 \)[/tex] molecules of [tex]\( O_2 \)[/tex]:
[tex]\[ C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O \][/tex]
The correctly balanced equation is:
[tex]\[ C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O \][/tex]
Thus, the sequence of coefficients needed to balance this chemical equation is:
[tex]\[ \boxed{1, 6, 6, 6} \][/tex]
