Respuesta :
Answer:
For a: The theoretical yield of carbon dioxide is 9.28 grams.
For b: The percent yield of the reaction is 72.2 %.
Explanation:
- For a:
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 carbon monoxide = 5.91 g
Molar mass of carbon monoxide = 28 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of carbon monoxide}=\frac{5.91g}{28g/mol}=0.211mol[/tex]
The chemical equation for the reaction of carbon monoxide and oxygen gas follows:
[tex]2CO(g)+O_2(g)\rightarrow 2CO_2(g)[/tex]
By Stoichiometry of the reaction:
2 moles of carbon monoxide produces 2 moles of carbon dioxide
So, 0.211 moles of carbon monoxide will produce = [tex]\frac{2}{2}\times 0.211=0.211mol[/tex] of carbon dioxide
Now, calculating the mass of carbon dioxide by using equation 1, we get:
Molar mass of carbon dioxide = 44 g/mol
Moles of carbon dioxide = 0.211 mol
Putting values in equation 1, we get:
[tex]0.211mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.211mol\times 44g/mol)=9.28g[/tex]
Hence, the theoretical yield of carbon dioxide is 9.28 grams.
- For b:
To calculate the percentage yield of carbon dioxide, we use the equation:
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Experimental yield of carbon dioxide = 6.70 g
Theoretical yield of carbon dioxide = 9.28 g
Putting values in above equation, we get:
[tex]\%\text{ yield of carbon dioxide}=\frac{6.70g}{9.28g}\times 100\\\\\% \text{yield of carbon dioxide}=72.2\%[/tex]
Hence, the percent yield of the reaction is 72.2 %.
