Respuesta :
Answer: The percent yield of the water is 31.98 %
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
- For methane:
Given mass of methane = 6.58 g
Molar mass of methane = 16 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of methane}=\frac{6.58g}{16g/mol}=0.411mol[/tex]
- For oxygen gas:
Given mass of oxygen gas = 14.4 g
Molar mass of oxygen gas = 32 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of oxygen gas}=\frac{14.4g}{32g/mol}=0.45mol[/tex]
The chemical equation for the combustion of methane is:
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]
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 = [tex]\frac{1}{2}\times 0.45=0.225mol[/tex] 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 = [tex]\frac{2}{2}\times 0.45=0.45[/tex] 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:
[tex]0.45mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.45mol\times 18g/mol)=8.1g[/tex]
- To calculate the percentage yield of water, we use the equation:
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Experimental yield of water = 2.59 g
Theoretical yield of water = 8.1 g
Putting values in above equation, we get:
[tex]\%\text{ yield of water}=\frac{2.59g}{8.1g}\times 100\\\\\% \text{yield of water}=31.98\%[/tex]
Hence, the percent yield of the water is 31.98 %
