A combustion reaction involves the release of heat from the burning of a hydrocarbon in oxygen to form carbon dioxide and water. The amount of hydrogen, carbon, and oxygen are conserved. This reaction is exothermic because of the high energy released when organic hydrocarbons are burned. It can be easily observed in the burning of petroleum distillates such as gasoline. 

There are five main types of chemical reactions.

1) Synthesis reactions have the following format:

$\displaystyle A + B \rightarrow AB$

In these reactions, two substances combine to form one substance. This occurs when hydrogen and oxygen combine to form water: 

$\displaystyle 2H_{2}(g) + O_{2} (g) \rightarrow 2H_{2}O (g)$

2) Decomposition reactions have the following format:

$\displaystyle AB \rightarrow A + B$

In these reactions, a compound breaks down to form smaller substances. This occurs when some acids decompose into an acidic oxide and water:

$\displaystyle H_{2}SO_{3} (aq) \rightarrow H_{2}O (aq) + SO_{2} (aq)$

3) Single displacement reactions have the following format:

$\displaystyle A + BC \rightarrow B + AC$

In these reactions, a single element replaces another element in a compound. This occurs in the reaction between zinc and hydrochloric acid:

$\displaystyle Zn (s) + HCl (aq) \rightarrow H_{2} (g) + ZnCl_{2} (aq)$

4) Double displacement reactions have the following format:

$\displaystyle AB + CD \rightarrow AC + BD$

In these reactions, an element from each of the two reactant compounds switches places. This occurs in the reaction between sulfuric acid and sodium hydroxide:

$\displaystyle H_{2}SO_{4} (aq) + 2NaOH (aq) \rightarrow Na_{2}SO_{4} (aq) + 2H_{2}O (aq)$

5) Combustion reactions have the following format:

$\displaystyle Hydrocarbon + Oxygen \rightarrow CO_{2} (g) + H_{2}O (g)$

In these reactions, a hydrocarbon and oxygen always react to form carbon dioxide and water. Here is an example:

$\displaystyle CH_{4} (g) + 2O_{2} (g) \rightarrow CO_{2}(g) + 2H_{2}O (g)$

$\displaystyle 4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)$

Recall that combustion equations are examples of reduction-oxidation reactions.

This cannot be a precipitation reaction because there are no solids in the products.

This cannot be a gas-evolution reaction because the reactants are not in aqueous form.

This cannot be an acid-base reaction because there is no proton transfer.

This is a basic identify the reaction type equation of the form:

$\displaystyle A+B\rightarrow AB_2$

Since both reactants are combined into one it is a combination reaction.

Now let's go over the wrong answers:

1) It isn't a single displacement because single displacement reactions are of the form:

$\displaystyle AB+C\rightarrow AC+B$

2) It isn't a double displacement/metathesis reaction because double displacement reactions are of the form:

$\displaystyle AB+CD\rightarrow AC+BD$

3) It isn't a decomposition reaction because decomposition reactions are the exact opposite of what occurs in the reaction in the problem. Meaning instead of two or more elements forming a compound, a compound breaks down into two or more elements.

$\displaystyle AB \rightarrow A+B$

4) It isn't a combustion reaction because a combustion reaction refers to the burning of a compound (usually a hydrocarbon) within which the other reactant is $\displaystyle O_2$ and the products are some amount of $\displaystyle CO_2$ + H_2$O$. So the general form of a combustion equation look as follows:

$\displaystyle C_x$H_y$ +O_2$\rightarrow CO_2$ + H_2O$

Combustion is the chemical reaction of a hydrocarbon with molecular oxygen, and it always produces carbon dioxide and water. Knowing the reactants and products, the unbalanced equation must be: 

$\displaystyle C_4H_{10} + O_2 \rightarrow CO_2 + H_2O$

We start by balancing the hydrogens. Since there are 10 on the left and only 2 on the right, we put a coefficient of 5 on water.

$\displaystyle C_4H_{10} + O_2 \rightarrow CO_2 + 5H_2O$

Similarly, we balance carbons by putting a 4 on the carbon dioxide.

$\displaystyle C_4H_{10} + O_2 \rightarrow 4CO_2 + 5H_2O$

To find the number of oxygens on the right, we multiply the 4 coefficient by the 2 subscript on O (which gets us 8 oxygens) and then add the 5 oxygens from the 5 water molecules to get a total of 13. The needed coefficient for $\displaystyle O_2$ on the left would then have to be 13/2.

$\displaystyle C_4H_{10} + \frac{13}{2}O_2 \rightarrow 4CO_2 + 5H_2O$

Because fractional coefficients are not allowed, we mutiply every coefficient by 2 to find our final reaction:

$\displaystyle 2C_4H_{10} + 2*\frac{13}{2}O_2 \rightarrow 2*4CO_2 + 2*5H_2O$

$\displaystyle 2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O$