Question

Please help !!

Answer 1

Answer:

For 2: The % yield of the product is 92.34 %

For 3: 12.208 L of carbon dioxide will be formed.

Explanation:

• For 2:

The percent yield of a reaction is calculated by using an equation:

$\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100$              ......(1)

Given values:

Actual value of the product = 78.4 g

Theoretical value of the product = 84.9 g

Plugging values in equation 1:

$\% \text{yield}=\frac{78.4 g}{84.9g}\times 100\\\\\% \text{yield}=92.34\%$

Hence, the % yield of the product is 92.34 %

• For 3:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(2)

Given mass of carbon dioxide = 24 g

Molar mass of carbon dioxide = 44 g/mol

Plugging values in equation 1:

$\text{Moles of carbon dioxide}=\frac{24g}{44g/mol}=0.545 mol$

At STP conditions:

1 mole of a gas occupies 22.4 L of volume

So, 0.545 moles of carbon dioxide will occupy = $\frac{22.4L}{1mol}\times 0.545mol=12.208L$ of volume

Hence, 12.208 L of carbon dioxide will be formed.