Answer: The percent yield of oxygen gas is 67.53 %.
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)
Given mass of water = 17.0 g
Molar mass of water = 18 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of water}=\frac{17.0g}{18g/mol}=0.944mol[/tex]
For the given chemical equation:
[tex]2H_2O\rightarrow 2H_2+O_2[/tex]
By Stoichiometry of the reaction:
2 moles of water produces 1 mole of oxygen gas
So, 0.944 moles of water will produce = [tex]\frac{1}{2}\times 0.944=0.472moles[/tex] of oxygen gas
Now, calculating the mass of oxygen gas from equation 1, we get:
Molar mass of oxygen gas = 32 g/mol
Moles of oxygen gas = 0.472 moles
Putting values in equation 1, we get:
[tex]0.472mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(0.472mol\times 32g/mol)=15.104g[/tex]
To calculate the percentage yield of oxygen gas, we use the equation:
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Experimental yield of oxygen gas = 10.2 g
Theoretical yield of oxygen gas = 15.104 g
Putting values in above equation, we get:
[tex]\%\text{ yield of oxygen gas}=\frac{10.2g}{15.104g}\times 100\\\\\% \text{yield of oxygen gas}=67.53\%[/tex]
Hence, the percent yield of oxygen gas is 67.53 %.
