Question

balance the reaction, do not leave any fractions, dont leave anything blank, then answer the questions about the reaction
1c₆h₆ +2o₂ →3co₂ + 4h₂o
this is an 5 reaction. is this a redox reaction? (yes/no)6, because c⁻⁴ is oxidized to c⁺⁴, and o₂⁰ is 7 to o⁻².
will the reaction happen as it is written? (yes/no)8, because it is a known exothermic reaction and because the products are stable. the reverse reaction would not happen.
a. 1 b. 2 c. 3 d. 4 e. 5 f. 6 g. 7 h. 8
i. 9 j. 10 k. 12 l. 14 m. 15 n. 18 o. 20 p. 21
q. 24 r. 26 s. 28 t. synthesis u. decomposition
v. single displacement w. double displacement x. neutralization
y. combustion z. yes aa. no bb. reactive cc. stable
dd. oxidized ee. reduced

Explanation:

Step1: Balance the carbon atoms

For \(C_6H_6\), there are 6 carbon atoms. So if we have \(x\) moles of \(C_6H_6\), \(y\) moles of \(O_2\), \(z\) moles of \(CO_2\) and \(w\) moles of \(H_2O\). From carbon balance: \(6x = z\). Let \(x = 2\), then \(z=12\).

Step2: Balance the hydrogen atoms

From hydrogen balance in \(C_6H_6\) and \(H_2O\): \(6x=2w\). Since \(x = 2\), then \(w = 6\).

Step3: Balance the oxygen atoms

From oxygen balance in \(O_2\), \(CO_2\) and \(H_2O\): \(2y=2z + w\). Substitute \(z = 12\) and \(w=6\), we get \(2y=2\times12 + 6=30\), so \(y = 15\).

The balanced equation is \(2C_6H_6+15O_2
ightarrow12CO_2 + 6H_2O\)

For the type of reaction:

• Combustion reaction: Hydrocarbons react with oxygen to produce carbon dioxide and water. \(C_6H_6\) (a hydrocarbon) reacts with \(O_2\) to form \(CO_2\) and \(H_2O\), so it is a combustion reaction (Y).
• Redox reaction: In \(C_6H_6\), carbon has an oxidation state of \(- 1\) (calculated as \(6x+6 = 0\), \(x=-1\)), in \(CO_2\) carbon has an oxidation state of \(+4\) (oxidized). In \(O_2\) oxygen has an oxidation state of \(0\), in \(CO_2\) and \(H_2O\) oxygen has an oxidation state of \(-2\) (reduced). So it is a redox reaction (Z).
• The reaction will happen as written (Z) because it is a combustion (exothermic) reaction and products (\(CO_2\) and \(H_2O\)) are stable.

Answer:

1. \(2\)
2. \(15\)
3. \(12\)
4. \(6\)
5. \(Y\) (combustion)
6. \(Z\) (yes)
7. \(EE\) (reduced)
8. \(Z\) (yes)