Chapter 3, Problem 157QRT

(a)

Interpretation Introduction

Interpretation:

$100.0g$ cobalt and $200.0g$ carbon monoxide gives $\text{89}\text{.7 \%}$ yield of dicobalt octacarbonyl (${\text{Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}$), the mass (g) of dicobalt octacarbonyl has to be calculated.

Explanation of Solution

Given,

  $100.0g$ cobalt and $200.0g$ carbon monoxide gives $\text{89}\text{.7 \%}$ yield of dicobalt octacarbonyl (${\text{Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}$)

Molar Mass of $\text{CO}$ is $\text{28}\text{.0101 g/mol}$, molar Mass of ${\text{Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}$ is $\text{341}\text{.9472 g/mol}$.

Hence, $\text{1 lb = 453}\text{.592}$

Balanced equation is given below,

  $\text{2Co}+8\text{CO}\to {\text{ Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}$

Two moles of cobalt reacts with eight moles of carbon monoxide gives one mole of dicobalt octacarbonyl.

Mass in grams of dicobalt octacarbonyl is calculated as follows,

$\begin{array}{l}\text{Mass}\text{of}{\text{Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}\text{ = 100}g\text{Co×}\frac{\text{1}\text{mol}\text{Co}}{\text{58}\text{.9332}\text{g}\text{Co}}\text{×}\frac{1\text{mol}{\text{Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}}{2\text{mol}\text{Co}}\text{×}\frac{\text{341}\text{.9472}\text{g}{\text{Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}}{{\text{1 mol Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}}\\ \text{Mass}\text{of}{\text{Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}\text{= 290}\text{.1 g}\end{array}$

$\begin{array}{l}\text{Mass}\text{of}{\text{Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}\text{ = 200}g\text{Co×}\frac{\text{1}\text{mol}\text{CO}}{\text{28}\text{.0101}\text{g}\text{CO}}\text{×}\frac{1\text{mol}{\text{Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}}{8\text{mol}\text{CO}}\text{×}\frac{\text{341}\text{.9472}\text{g}{\text{Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}}{{\text{1 mol Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}}\\ \text{Mass}\text{of}{\text{Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}\text{= 305}\text{.1 g}\end{array}$

According to the gram calculation, cobalt is limiting reagent, therefore the gram of cobalt is calculated as follows,

    $\begin{array}{l}\text{Grams}\text{of}\text{Co}\text{= 290}\text{.1}\text{g}\text{×}\frac{\text{89}\text{.7}\text{g}\text{actaul}\text{yield}}{\text{100}\text{g}\text{theoritical}\text{yield}}\\ \text{Grams}\text{of}\text{Co}\text{= 260}\text{.2 g}\end{array}$

The mass (g) of dicobalt octacarbonyl is $\text{260}\text{.2 g}$.

(b)

Interpretation Introduction

Interpretation:

The additional mass of the limiting reactant that is needed to react completely with the residual excess reactant has to be calculated.

Explanation of Solution

Given,

  $100.0g$ cobalt and $200.0g$ carbon monoxide gives $\text{89}\text{.7 \%}$ yield of dicobalt octacarbonyl (${\text{Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}$)

Molar Mass of $\text{CO}$ is $\text{28}\text{.0101 g/mol}$, molar Mass of ${\text{Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}$ is $\text{341}\text{.9472 g/mol}$.

Assuming that the carbon monoxide in the final product is the only carbon monoxide used.

Mass in grams of dicobalt octacarbonyl is calculated as follows,

$\begin{array}{l}\text{Mass}\text{of}\text{CO used = 260}g{\text{Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}\text{×}\frac{\text{1}\text{mol}{\text{Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}}{\text{341}\text{.9472}\text{g}{\text{Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}}\text{×}\frac{8\text{mol}\text{CO}}{1\text{mol}{\text{Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}}\text{×}\frac{\text{28}\text{.0101}\text{g}\text{CO}}{\text{1 mol CO}}\\ \text{Mass}\text{of}\text{CO used}\text{= 170}\text{.38 g}\end{array}$

Excess CO is calculated as follows,

  $\begin{array}{l}\text{Excess}\text{CO = 200 g CO initial -}\text{170}\text{.38 g CO used}\\ \text{Excess}\text{CO = 29}\text{.62 g}\end{array}$

Balanced equation is given below,

  $\text{2Co}+8\text{CO}\to {\text{ Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}$

Two moles of cobalt reacts with eight moles of carbon monoxide gives one mole of dicobalt octacarbonyl.

Grams of cobalt is calculated for reaction of excess carbon monoxide is given below,

$\begin{array}{l}\text{Grams of}\text{Co = 29}\text{.62}g\text{CO ×}\frac{\text{1}\text{mol}\text{CO}}{\text{28}\text{.0101}\text{g}\text{CO}}\text{×}\frac{2\text{mol}\text{Co}}{8\text{mol}\text{CO}}\text{×}\frac{\text{58}\text{.9332}\text{g}\text{Co}}{\text{1 mol Co}}\\ \text{Grams of}\text{Co}\text{= 15}\text{.58 g}\end{array}$

Therefore, $\text{15}\text{.58 g}$ of cobalt reacts with excess carbon monoxide.

(c)

Interpretation Introduction

Interpretation:

Total mass of dicobalt octacarbonyl produced from subpart (a) and (b) has as to be calculated.

Explanation of Solution

Given,

  $100.0g$ cobalt and $200.0g$ carbon monoxide gives $\text{89}\text{.7 \%}$ yield of dicobalt octacarbonyl (${\text{Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}$)

Molar Mass of $\text{CO}$ is $\text{28}\text{.0101 g/mol}$, molar Mass of ${\text{Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}$ is $\text{341}\text{.9472 g/mol}$.

Balanced equation is given below,

  $\text{2Co}+8\text{CO}\to {\text{ Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}$

Two moles of cobalt reacts with eight moles of carbon monoxide gives one mole of dicobalt octacarbonyl.

Mass in grams of dicobalt octacarbonyl is calculated as follows,

$\begin{array}{l}\text{Mass}\text{of}{\text{Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}\text{ = 200}g\text{Co×}\frac{\text{1}\text{mol}\text{CO}}{\text{28}\text{.0101}\text{g}\text{CO}}\text{×}\frac{1\text{mol}{\text{Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}}{8\text{mol}\text{CO}}\text{×}\frac{\text{341}\text{.9472}\text{g}{\text{Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}}{{\text{1 mol Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}}\\ \text{Mass}\text{of}{\text{Co}}_{\text{2}}{\left(\text{CO}\right)}_{\text{8}}\text{= 305}\text{.1 g}\end{array}$

Therefore the total gram of dicobalt octacarbonyl is calculated as follows,

    $\begin{array}{l}\text{Mass}\text{of}{\text{Co(CO}}_{\text{8}}\text{) = 305}\text{.1}\text{g}\text{×}\frac{\text{89}\text{.7}\text{g}\text{actaul}\text{yield}}{\text{100}\text{g}\text{theoritical}\text{yield}}\text{= 273}\text{.6 g}\\ \text{Mass}\text{of}{\text{Co(CO}}_{\text{8}}\text{)}\text{= 274 g}\end{array}$

Total mass of dicobalt octacarbonyl produced is $\text{274 g}$.