Question 4.8: Balancing a Chemical Equation . Balance this equation for th...

Balancing a Chemical Equation . Balance this equation for the combustion of butane, the fluid most commonly used in pocket lighters:

Balancing a Chemical Equation . Balance this equation for the combustion of butane, the fluid most commonly used in pocket lighters:Strategy : The equation for the combustion of butane is very similar to the one we examined at the beginning of this section for the combustion of propane. To

Strategy : The equation for the combustion of butane is very similar to the one we examined at the beginning of this section for the combustion of propane. To balance an equation, we place numbers in front of the formulas until there are identical numbers of atoms on each side of the equation.

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To balance carbons, put a 4 in front of the CO_{2} (because there are four carbons on the left). Then to balance hydrogens, place a 5 in front of the H_{2} O to give ten hydrogens on each side of the equation.

C_{4} H_{10} (g)+O_{2} (g) \rightarrow 4CO_{2} (g)+ 5H_{2} O (g)

When we count the oxygens, we find 2 on the left and 13 on the right. We can balance their numbers by putting 13/2 in front of the O_{2}.

C_{4} H_{10} (g)+ \frac{13}{2} O_{2} (g) \rightarrow 4CO_{2} (g)+ 5H_{2} O (g)

Although chemists sometimes have good reason to write equations with fractional coefficients, it is common practice to use only whole-number coefficients. We accomplish that by multiplying everything by 2, which gives the balanced equation:

2C_{4} H_{10} (g)+ 13 O_{2} (g) \rightarrow 8CO_{2} (g)+ 10H_{2} O (g)

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