The world burns the fossil fuel equivalent of approximately $9.50\times 10^{12}$ kg of petroleum per year. Assume that all of this petroleum is in the form of octane. Calculate how much CO2 in kilograms is produced by world fossil fuel combustion per year.( Hint: Begin by writing a balanced equation for the combustion of octane.)

Answer: $29\times 10^{12}kg$

Explanation:

Combustion is a chemical reaction in which hydrocarbons are burnt in the presence of oxygen to give carbon dioxide and water.

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

$2_C8H_{18}+17O_2\rightarrow 16CO_2+18H_2O$

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of octane}=\frac{9.50\times 10^{12}\times 10^3g}{114g/mol}=0.083\times 10^{15}moles$

According to stoichiometry :

As 2 moles of octane give = 16 moles of $CO_2$

Thus $0.083\times 10^{15}moles$ of octane give = $\frac{16}{2}\times 0.083\times 10^{15}=0.664\times 10^{15}moles$ of $CO_2$

Mass of $CO_2=moles\times {\text {Molar mass}}=0.664\times 10^{15}moles\times 44g/mol=29.2\times 10^{15}g=29.2\times 10^{12}kg$

Thus $29\times 10^{12}kg$ of $CO_2$ is produced by world fossil fuel combustion per year.

5 0

3 years ago

In a chemical reaction, what is defined as the difference between the potential energy of the products and the potential energy

zlopas [31]

I Believe That The Answer Is A
Hope This Helps !

6 0

2 years ago

Read 2 more answers

This is the chemical formula for talc (the main ingredient in talcum powder): mg3(si2o5)2(oh)2. an analytical chemist fins that

weqwewe [10]

In one mole of talc, we observe that there are:
3 moles of Mg
4 moles of Si
2 moles of H
12 moles of O

The molar ratio of O to Mg is then:

12 moles of O : 3 moles of Mg = 4 : 1

Therefore, if 6.1 moles of Mg are present, the moles of O are:

4 * 6.1 = 24.4 moles of O

7 0

3 years ago