Answer:
For 2: The % yield of the product is 92.34 %
For 3: 12.208 L of carbon dioxide will be formed.
Explanation:
The percent yield of a reaction is calculated by using an equation:
[tex]\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100[/tex] ......(1)
Given values:
Actual value of the product = 78.4 g
Theoretical value of the product = 84.9 g
Plugging values in equation 1:
[tex]\% \text{yield}=\frac{78.4 g}{84.9g}\times 100\\\\\% \text{yield}=92.34\%[/tex]
Hence, the % yield of the product is 92.34 %
The number of moles is defined as the ratio of the mass of a substance to its molar mass.
The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(2)
Given mass of carbon dioxide = 24 g
Molar mass of carbon dioxide = 44 g/mol
Plugging values in equation 1:
[tex]\text{Moles of carbon dioxide}=\frac{24g}{44g/mol}=0.545 mol[/tex]
At STP conditions:
1 mole of a gas occupies 22.4 L of volume
So, 0.545 moles of carbon dioxide will occupy = [tex]\frac{22.4L}{1mol}\times 0.545mol=12.208L[/tex] of volume
Hence, 12.208 L of carbon dioxide will be formed.
