Question

homework: chapter 6
exercise 6.99
the citizens of the world burn the fossil - fuel equivalent of 2.0×10¹² kg of petroleum per year. assume that all of this petroleum is in the form of nonane (c₉h₂₀).
part a
calculate how much co₂ (in kg) is produced by world fossil - fuel combustion per year. (hint: begin by writing a balanced equation for the combustion of nonane.) when calculating this answer, do not round any values in your calculations until the end and when you are reporting the final value.
express your answer in kilograms to two significant figures.
mco₂ = kg

Explanation:

Step1: Write balanced combustion equation

The balanced combustion equation for nonane ($C_9H_{20}$) is $C_9H_{20}+14O_2
ightarrow9CO_2 + 10H_2O$.

Step2: Calculate moles of nonane

The molar mass of nonane ($C_9H_{20}$) is $M_{C_9H_{20}}=9\times12.01+20\times1.01 = 128.29\ g/mol=0.12829\ kg/mol$. The mass of nonane burned per - year is $m = 2.0\times10^{14}\ kg$. The number of moles of nonane, $n_{C_9H_{20}}=\frac{m}{M}=\frac{2.0\times10^{14}\ kg}{0.12829\ kg/mol}\approx1.56\times10^{15}\ mol$.

Step3: Determine moles of $CO_2$

From the balanced equation, the mole - ratio of $C_9H_{20}$ to $CO_2$ is $1:9$. So the number of moles of $CO_2$ produced, $n_{CO_2}=9\times n_{C_9H_{20}}=9\times1.56\times10^{15}\ mol = 1.404\times10^{16}\ mol$.

Step4: Calculate mass of $CO_2$

The molar mass of $CO_2$ is $M_{CO_2}=12.01 + 2\times16.00=44.01\ g/mol = 0.04401\ kg/mol$. The mass of $CO_2$ produced, $m_{CO_2}=n_{CO_2}\times M_{CO_2}=1.404\times10^{16}\ mol\times0.04401\ kg/mol\approx6.2\times10^{14}\ kg$.

Answer:

$6.2\times10^{14}\ kg$