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german
4 years ago
10

Assume that all of this fossil fuel is in the form of octane (C8H18) and calculate how much CO2 in kilograms is produced by worl

d fossil fuel combustion per year. (Hint: Begin by writing a balanced equation for the combustion of octane.) Express your answer using two significant figures.
Chemistry
1 answer:
AnnZ [28]4 years ago
5 0

The given question is incomplete. the complete question is:

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.

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If mercury (Hg) and oxygen (O2) were reacted to form mercury oxide how many molecules of each reactant and product would be need
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3 years ago
There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with
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\text{Hg} \text{O} and \text{Hg}_{2} \text{O}.

Assuming complete decomposition of both samples,

  • m(\text{Hg}) = m(\text{residure})
  • m(\text{O}) = m(\text{loss})

First compound:

  • m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g

n = m/M; 0.6498 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.048 \; g  / 16 \; g \cdot mol^{-1}= 0.003 \; mol
  • n(\text{Hg atoms}) = 0.6018 \; g  / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol

Oxygen and mercury atoms seemingly exist in the first compound at a 1:1 ratio; thus the empirical formula for this compound would be \text{Hg} \text{O} where the subscript "1" is omitted.

Similarly, for the second compound

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n = m/M; 0.4172 \; g of the first compound would contain

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n(\text{Hg}) : n(\text{O}) \approx  2:1 and therefore the empirical formula

\text{Hg}_{2} \text{O}.

8 0
3 years ago
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timofeeve [1]

Answer:

Answer:

step 1:balance skeleton equation the chemical equation:

Zn +HNO3➔Zn(NO3)2+NO+H2O

step 2: identity undergoing oxidation or reduction

here

Zn➔Zn(NO3)2

Zn is oxidized from 0 to 2 in oxidation no.

HNO3➔NO

N is reduced from 5 to 2 in oxidation no

Step 3: calculate change in oxidation no.

change in oxidation no

in Zn=0-2=-2=2

in

N=5-2=3

Step 4: Balance it by doing crisscrossed multiplication

we get;

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step 6:Balance other atoms except H & O

3Zn +2HNO3➔3Zn(NO3)2+2NO+H2O

3Zn +2HNO3+6HNO3➔3Zn(NO3)2+2NO+H2O

finally: balance H

<em><u>3Zn +8HNO3➔3Zn(NO3)2+2NO+4H2O</u></em>

3 0
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Read 2 more answers
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