<u>Answer:</u> The percent yield of the water is 31.98 %
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
Given mass of methane = 6.58 g
Molar mass of methane = 16 g/mol
Putting values in equation 1, we get:
![\text{Moles of methane}=\frac{6.58g}{16g/mol}=0.411mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20methane%7D%3D%5Cfrac%7B6.58g%7D%7B16g%2Fmol%7D%3D0.411mol)
Given mass of oxygen gas = 14.4 g
Molar mass of oxygen gas = 32 g/mol
Putting values in equation 1, we get:
![\text{Moles of oxygen gas}=\frac{14.4g}{32g/mol}=0.45mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20oxygen%20gas%7D%3D%5Cfrac%7B14.4g%7D%7B32g%2Fmol%7D%3D0.45mol)
The chemical equation for the combustion of methane is:
![CH_4+2O_2\rightarrow CO_2+2H_2O](https://tex.z-dn.net/?f=CH_4%2B2O_2%5Crightarrow%20CO_2%2B2H_2O)
By Stoichiometry of the reaction:
2 moles of oxygen gas reacts with 1 mole of methane
So, 0.45 moles of oxygen gas will react with =
of methane
As, given amount of methane is more than the required amount. So, it is considered as an excess reagent.
Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction
2 moles of oxygen gas produces 2 moles of water
So, 0.45 moles of oxygen gas will produce =
moles of water
- Now, calculating the mass of water from equation 1, we get:
Molar mass of water = 18 g/mol
Moles of water = 0.45 moles
Putting values in equation 1, we get:
![0.45mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.45mol\times 18g/mol)=8.1g](https://tex.z-dn.net/?f=0.45mol%3D%5Cfrac%7B%5Ctext%7BMass%20of%20water%7D%7D%7B18g%2Fmol%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20water%7D%3D%280.45mol%5Ctimes%2018g%2Fmol%29%3D8.1g)
- To calculate the percentage yield of water, we use the equation:
![\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100](https://tex.z-dn.net/?f=%5C%25%5Ctext%7B%20yield%7D%3D%5Cfrac%7B%5Ctext%7BExperimental%20yield%7D%7D%7B%5Ctext%7BTheoretical%20yield%7D%7D%5Ctimes%20100)
Experimental yield of water = 2.59 g
Theoretical yield of water = 8.1 g
Putting values in above equation, we get:
![\%\text{ yield of water}=\frac{2.59g}{8.1g}\times 100\\\\\% \text{yield of water}=31.98\%](https://tex.z-dn.net/?f=%5C%25%5Ctext%7B%20yield%20of%20water%7D%3D%5Cfrac%7B2.59g%7D%7B8.1g%7D%5Ctimes%20100%5C%5C%5C%5C%5C%25%20%5Ctext%7Byield%20of%20water%7D%3D31.98%5C%25)
Hence, the percent yield of the water is 31.98 %