Questions: A reaction below produced nickel tetracarbonyl, Ni(CO)4 in a 88% yield. Assuming sufficient nickel metal was used, what mass of carbon monoxide, CO, was added to the reaction if the actual yield of product, Ni(CO)4, was 185 g? Hint: Calculate the theoretical yield of nickel tetracarbonyl from the percent yield and work backwards from there. 4 CO(g)+Ni(s) -> Ni(CO)4(l) - For CO, 28.0 g / mol - For Ni(CO)4, 170.7 g / mol

A reaction below produced nickel tetracarbonyl, Ni(CO)4 in a 88% yield. Assuming sufficient nickel metal was used, what mass of carbon monoxide, CO, was added to the reaction if the actual yield of product, Ni(CO)4, was 185 g? Hint: Calculate the theoretical yield of nickel tetracarbonyl from the percent yield and work backwards from there.

4 CO(g)+Ni(s) -> Ni(CO)4(l)

- For CO, 28.0 g / mol
- For Ni(CO)4, 170.7 g / mol

Transcript text: A reaction below produced nickel tetracarbonyl, $\mathrm{Ni}(\mathrm{CO})_{4}$ in a $88 \%$ yield. Assuming sufficient nickel metal was used, what mass of carbon monoxide, CO , was added to the reaction if the actual yield of product, $\mathrm{Ni}(\mathrm{CO}) 4$, was 185 g ? Hint: Calculate the theoretical yield of nickel tetracarbonyl from the percent yield and work backwards from there. \[ 4 \mathrm{CO}(\mathrm{~g})+\mathrm{Ni}(\mathrm{~s}) \rightarrow \mathrm{Ni}(\mathrm{CO})_{4}(\mathrm{l}) \] - For CO, $28.0 \mathrm{~g} / \mathrm{mol}$ - For $\mathrm{Ni}(\mathrm{CO})_{4} .170 .7 \mathrm{~g} / \mathrm{mol}$

Solution

Solution Steps

Step 1: Calculate the Theoretical Yield of Ni(CO)$_4$

Given that the actual yield of Ni(CO)$_4$ is 185 g and the percent yield is 88%, we can calculate the theoretical yield using the formula:

\[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% \]

Rearranging to find the theoretical yield:

\[ \text{Theoretical Yield} = \frac{\text{Actual Yield}}{\text{Percent Yield}} \times 100\% \]

Substituting the given values:

\[ \text{Theoretical Yield} = \frac{185 \, \text{g}}{88\%} \times 100\% = \frac{185}{0.88} \approx 210.2273 \, \text{g} \]

Step 2: Calculate Moles of Ni(CO)$_4$ Produced

Using the molar mass of Ni(CO)$_4$ (170.7 g/mol), calculate the moles of Ni(CO)$_4$ produced:

\[ \text{Moles of Ni(CO)}_4 = \frac{\text{Theoretical Yield}}{\text{Molar Mass of Ni(CO)}_4} = \frac{210.2273 \, \text{g}}{170.7 \, \text{g/mol}} \approx 1.2317 \, \text{mol} \]

Step 3: Calculate Moles of CO Required

From the balanced chemical equation, 4 moles of CO are required to produce 1 mole of Ni(CO)$_4$. Therefore, the moles of CO required are:

\[ \text{Moles of CO} = 4 \times \text{Moles of Ni(CO)}_4 = 4 \times 1.2317 \, \text{mol} \approx 4.9268 \, \text{mol} \]

Step 4: Calculate Mass of CO Added

Using the molar mass of CO (28.0 g/mol), calculate the mass of CO added:

\[ \text{Mass of CO} = \text{Moles of CO} \times \text{Molar Mass of CO} = 4.9268 \, \text{mol} \times 28.0 \, \text{g/mol} \approx 137.9504 \, \text{g} \]