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balance the reaction, do not leave any fractions, dont leave anything b…

Question

balance the reaction, do not leave any fractions, dont leave anything blank, then answer the questions about the reaction
1 c₉h₁₈ + 2 o₂ → 3 co₂(g) + 4 h₂o
this is an 5 reaction.
is this a redox reaction? (yes/no)6, because o₂⁰ is reduced to o⁻², and because c⁻⁴ is 7 to c⁺⁴.
will the reaction happen as it is written? (yes/no)8, because this is known to be an exothermic reaction and because co₂ and h₂o are 9. 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. 16 n. 18 o. 20 p. 21
q. 24 r. 26 s. 27 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 carbon atoms

For \(C_9H_{18}\), there are 9 carbon atoms. So, the coefficient of \(CO_2\) (\(3\)) is 9.

Step2: Balance hydrogen atoms

There are 18 hydrogen atoms in \(C_9H_{18}\). So, the coefficient of \(H_2O\) (\(4\)) is 9.

Step3: Balance oxygen atoms

The right - hand side has \(9\times2 + 9\times1=27\) oxygen atoms. So, the coefficient of \(O_2\) (\(2\)) is \(\frac{27}{2}\). But we need to multiply all coefficients by 2 to get rid of the fraction. So, the balanced equation is \(2C_9H_{18}+27O_2
ightarrow18CO_2 + 18H_2O\).

This is a combustion reaction (\(5 = Y\)).

Since \(O\) is reduced (\(O_2^0
ightarrow O^{- 2}\)) and \(C\) is oxidized (\(C^{-4}
ightarrow C^{+4}\)), it is a redox reaction (\(6 = Z\), \(7 = DD\)).

Combustion reactions are exothermic and produce stable products (\(CO_2\) and \(H_2O\) are stable, \(9 = CC\)). So, the reaction will happen as written (\(8 = Z\)).

Answer:

  1. \(2\)
  2. \(27\)
  3. \(18\)
  4. \(18\)
  5. \(Y\)
  6. \(Z\)
  7. \(DD\)
  8. \(Z\)
  9. \(CC\)