Question

How do you determine the gram-formula mass of the propane gas?State the mole ratio represented in the equation of oxygen to carbon dioxide.Calculate the number of liters of water vapor produced when 25.0 liters of oxygen gas are consumed?
The burning of propane gas can be represented as a balance chemical reaction as follows:
$C_3{H}_{8(g)}+5O_{2_{(g)}}\to 3C{O_2}_{(g)}+4H_2{O}_{\left(g\right)}$

Answer 1

Hint The mole ratio can be said as the ratio of the coefficients of two atoms in a balanced equation. Here, the gram –formula mass is the summation of the atomic mass of the atoms present in a compound.

Complete step by step solution:
The question asked above have three parts in them, where the first part is about finding the gram molecular formulae of the propane gas, next we have to find the mole ratio between the oxygen and carbon dioxide in the given chemical equation and the final part is to calculate the number of liters of water vapour formed when a volume of 25 L of oxygen gas is consumed.
Let's solve one by one, now the first part is all about the gram-formula mass, we are familiar with this term and the another term used for gram formula mass is generally said as the molecular mass of the compound but more accurately it is the amount of the substance in grams which is equal to the mass as per the formulae in atomic mass units.
Let’s calculate the gram formula mass of propane which has the formulae of $C_3{H}_8$ and here we have only one mole of $C_3{H}_8$, so just add up the atomic mass of the atoms in the formulae.
There are three carbon atoms and eight H atoms in the formula, the atomic mass of C is 12 and the atomic mass of H is 1.
So let’s calculate the gram formula mass of $C_3{H}_8$,
$\text{Gram}\text{formulae}\text{mass=3 }\times \left(\text{12}\right)\text{+8 }\times \left(\text{1}\right)\text{=44g/mol}$
So the first part is solved, now let’s move to the second section.
Here we have to write the mole ratio between oxygen and carbon dioxide in the given balanced equation of combustion of propane.
The mole ratio of two substances can be said as the ratio between the coefficients of the substances of concern in the balanced chemical equation.
Here we have to find the ratio of $O_2:C{O}_2$ as per given in the question. First observe the coefficients of these two substances in the balanced chemical equation.
The coefficient of $O_2$is 5 and the coefficient of $C{O}_2$ is 3.
Therefore the ratio is $5:3$.
5 moles of oxygen is required to produce 3 moles of carbon dioxide.
The third part of the question is to calculate the number of liters of water vapor produced when 25.0 liters of oxygen gas are consumed,for that we have to first find the number of moles of oxygen in 25L.
One mole of a gaseous substance at STP has a volume of about 22.4 l.
Now let's find the number of moles of $O_2$ in 25L
$\text{No}\text{.of moles of oxygen =}{\displaystyle \frac{\text{25L}}{\text{22}\text{.4L}}}\text{=1}\text{.12}\text{mole}\text{of}\text{oxygen}$
Now from this we have to find the moles of water and we use the equation,
$\text{No}\text{.of moles of water= No}\text{.of moles of water }\times {\displaystyle \frac{\text{4mol}\text{water}}{\text{5}\text{mol}\text{oxygen}}}$
$\text{No}\text{.of moles of water=1}\text{.12 }\times {\displaystyle \frac{\text{4}}{\text{5}}}\text{=0}\text{.896mol}\text{of}\text{water}$
Now we got the number of moles of water as in the question is asked about water vapour and the water vapour is the gaseous form of water and to get the volume of water vapour multiply the number of moles with one mole i.e. 22.4L
$\text{Volume}\text{of}\text{water}\text{vapour=0}\text{.896mol }\times \text{ 22}\text{.4L=20L}$

Note: If we have to calculate for water in liquid state then we will first find the weight of water by multiplying the number of moles obtained with the molecular mass of water and then give the mass obtained in the density equation to get the final volume.
The molecular mass of water is 18g
$\text{Weight}\text{of}\text{water=No}\text{.}\text{of}\text{moles}\text{of}\text{water }\times \text{ Molecular}\text{mass}\text{of}\text{water}$
$\text{Weight}\text{of}\text{water=0}\text{.896 }\times \text{ 18=16}\text{.128g}$
The density formula is,
$\text{density=}{\displaystyle \frac{\text{mass}}{\text{volume}}}$
The value of density of water is $1g/mL$
$\text{Volume=}{\displaystyle \frac{\text{mass}}{\text{density}}}\text{=}{\displaystyle \frac{\text{16}\text{.128}}{\text{1}}}\text{=16}\text{.128L}$ I.e. 16.128L of liquid water