if 0.45 grams of water is formed during the cumbustion of propane, how many grams of carbon dioxide is formed?

The stoichiometry of this reaction shows that every mole of propane reacts with 5 moles of oxygen to be burned and produce 3 moles of CO2 and 4 moles of water.

This mains that every 3 moles of CO2 produced with the production of 4 moles of water.

In the problem 0.45 g of water is formed. We can get the number of moles through the relation n = mass/molar mass, mass = 0.45 g and molar mass of water = 18.0 g/mol. n = 0.45/18 = 0.025 mole.

As we mentioned that every 3 moles of CO2 produced with 4 moles of water, then we can get the number of moles of CO2 when 0.025 mole of water formed.

The number of moles of CO2 corresponding to the formation of 0.025 mole of water = (0.025 * 3) / (4) = 0.01875 mole.

Finally we can get the number of grams of CO2 formed; mass = n * molar mass, molar mass of CO2 = 44.0 g/mole.

Mass of CO2 = (0.01875 mole * 44.0 g/mole) = 0.825 g.

aliasger2709 aliasger2709 · Answer 2 · 2022-03-21T08:09:18+01:00

A combustion reaction is the exothermic redox reaction of the reactants in the presence of oxygen. The mass of the carbon dioxide formed is 0.825 g.

What is mass?

Mass is the weight of the substance required in the chemical reaction. The combustion reaction of propane can be shown as:

[tex]\rm C_{3}H_{8} + 5O_{2} \rightarrow 3CO_{2} + 4H_{2}O[/tex]

From the reaction stoichiometry, one mole of propane reacts with 5 moles of oxygen molecules to produce 3 moles of carbon dioxide and 4 moles of water molecules.

From this, it can be said that 3 moles of carbon dioxide are produced along 4 moles of water. The moles of water can be calculated as:

[tex]\begin{aligned}\rm n &= \rm \dfrac{mass}{molar\;mass}\\\\&= \dfrac{0.45}{18}\\\\&=0.025\;\rm mol\end{aligned}[/tex]

The moles of carbon dioxide can be calculated from water as:

[tex]\dfrac{0.025\times 3}{4}= 0.01875\;\rm mol[/tex]

Mass of carbon dioxide can be calculated as:

[tex]\begin{aligned}\rm Mass&= \rm Molar \; mass \times moles \\\\&= 44\times 0.01875\\\\&=0.825 \;\rm gm\end{aligned}[/tex]

Therefore, 0.825 gm of carbon dioxide is formed.

Learn more about mass and moles here:

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