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Question: If 88g of \({C_3}{H_8}\) and 160g of \({O_2}\) are allowed to react to form \(C{O_2}\) and \({H_2}O\...

If 88g of C3H8{C_3}{H_8} and 160g of O2{O_2} are allowed to react to form CO2C{O_2} and H2O{H_2}O, how many grams of CO2C{O_2} will be formed.
A.33g
B.60g
C.132g
D.264g
E.None of the above

Explanation

Solution

We have to calculate the moles of carbon dioxide from propane and the moles of carbon dioxide from oxygen. From the calculated moles of carbon dioxide from each reagent, we have to predict the limiting reagent. The grams of carbon dioxide formed are calculated from moles of carbon dioxide of limiting reagent and the molar mass of carbon dioxide.

Complete step by step answer:
Given data contains,
Mass of C3H8{C_3}{H_8} is 88g.
Mass of O2{O_2} is 160g.
We obtain the products carbon dioxide and water from the combustion reaction of propane. We can write the balanced chemical equation as,
C3H8+5O23CO2+4H2O{C_3}{H_8} + 5{O_2} \to 3C{O_2} + 4{H_2}O
We know the molar mass of C3H8{C_3}{H_8} is 44.1g/mol.
We know the molar mass of O2{O_2} is 32g/mol.
We know the molar mass of CO2C{O_2} is 44g/mol.
Let us now calculate the moles of carbon dioxide from propane.
Moles of carbon dioxide = (88g×1mol44g)×(3molCO21molC3H8)\left( {88g \times \dfrac{{1mol}}{{44g}}} \right) \times \left( {\dfrac{{3molCO{ _2}}}{{1mol{C_3}{H_8}}}} \right)
Moles of carbon dioxide = 6moles6moles
The moles of carbon dioxide obtained from propane are six moles.
Let us now calculate the moles of carbon dioxide from oxygen.
Moles of carbon dioxide=(160g×1molO232g)×(3molCO25molO2)\left( {160g \times \dfrac{{1mol{O_2}}}{{32g}}} \right) \times \left( {\dfrac{{3molCO{ _2}}}{{5mol{O_2}}}} \right)
Moles of carbon dioxide = 3moles3moles
The moles of carbon dioxide obtained from oxygen are three moles.
From the calculated moles of carbon dioxide from oxygen and calculated moles of carbon dioxide from propane, we can see that oxygen is the limiting reagent.
So, we have to now calculate the grams of carbon dioxide from moles of carbon dioxide formed from oxygen and the molar mass of oxygen.
Grams of carbon dioxide = 3  moles  CO2×44g  CO2  1  mol  CO23\;moles\;C{O_2} \times \dfrac{{44g\;C{O_2}}\;}{{1\;mol\;C{O_2}}}
Grams of carbon dioxide = 132gCO2132gC{O_2}
The grams of carbon dioxide is 132g132g.

Therefore, the option (C) is correct.

Note:
As we know that the burning of fossil fuels by humans is the largest source of releases of carbon dioxide that is one of the greenhouse gases that permits radiative forcing and contributes to global warming. A small part of hydrocarbon-based fuels are biofuels obtained from atmospheric carbon dioxide, and therefore do not rise the net amount of carbon dioxide in the atmosphere. Combustion of fossil fuels produces sulfuric and nitric acids that fall to Earth as acid rain, affecting both natural areas and the built environment.