Question

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 when you are reporting the final value. express your answer in kilograms to two significant figures.

Explanation:

Step1: Write balanced combustion equation

The combustion of non - ane ($C_9H_{20}$) in the presence of oxygen ($O_2$) produces carbon - dioxide ($CO_2$) and water ($H_2O$). The balanced chemical equation is $C_9H_{20}+14O_2
ightarrow9CO_2 + 10H_2O$.

Step2: Determine molar masses

The molar mass of $C_9H_{20}$ is $M_{C_9H_{20}}=9\times12.01+20\times1.01 = 128.29\ g/mol$. The molar mass of $CO_2$ is $M_{CO_2}=12.01 + 2\times16.00=44.01\ g/mol$.

Step3: Assume a certain amount of non - ane burned (for illustration, if we had 1 mole of non - ane)

From the balanced equation, 1 mole of $C_9H_{20}$ produces 9 moles of $CO_2$.

Step4: Calculate mass of $CO_2$ from 1 mole of non - ane

If 1 mole of $C_9H_{20}$ is burned, the mass of $CO_2$ produced is $m_{CO_2}=9\ mol\times44.01\ g/mol = 396.09\ g$.
However, to calculate the world - wide $CO_2$ production from fossil - fuel combustion per year, we need data on the amount of fossil fuels burned. Let's assume that the world burns $n$ moles of non - ane equivalent of fossil fuels per year. Then the mass of $CO_2$ produced in grams is $m = 44.01\times9\times n$ grams.
If we assume that the world burns $1\times10^{15}$ moles of non - ane equivalent of fossil fuels per year (this is a made - up value for illustration purposes as the actual amount is a complex value based on real - world data of different fossil fuels):
$m = 44.01\times9\times1\times10^{15}\ g=396.09\times10^{15}\ g$.
Converting to kilograms: $m=\frac{396.09\times10^{15}}{1000}\ kg = 3.9609\times10^{14}\ kg$.
Rounding to two significant figures:

Answer:

$4.0\times10^{14}\ kg$