Respuesta :
Answer: The ideal yield of carbon dioxide is 7.506 grams
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 carbon monoxide = 5.61 g
Molar mass of carbon monoxide = 28 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of carbon monoxide}=\frac{5.61g}{28g/mol}=0.200mol[/tex]
The chemical equation for the reaction of carbon monoxide and water follows:
[tex]CO(g)+H_2O(l)\rightarrow CO_2(g)+H_2(g)[/tex]
By Stoichiometry of the reaction:
1 mole of carbon monoxide produces 1 mole of carbon dioxide
So, 0.200 moles of carbon monoxide will produce = [tex]\frac{1}{1}\times 0.200=0.200mol[/tex] of carbon dioxide
Now, calculating the mass of carbon dioxide from equation 1, we get:
Molar mass of carbon dioxide = 44 g/mol
Moles of carbon dioxide = 0.200 moles
Putting values in equation 1, we get:
[tex]0.200mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.200mol\times 44g/mol)=8.8g[/tex]
To calculate the experimental yield of carbon dioxide, we use the equation:
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Percentage yield of carbon dioxide = 85.3 %
Theoretical yield of carbon dioxide = 8.8 g
Putting values in above equation, we get:
[tex]85.3=\frac{\text{Experimental yield of carbon dioxide}}{8.8g}\times 100\\\\\text{Experimental yield of carbon dioxide}=\frac{85.3\times 8.8}{100}=7.506g[/tex]
Hence, the ideal yield of carbon dioxide is 7.506 grams
