Question

Calculate the volume of hydrogen at 0° C and 1.00 atm that is required to convert 500.0 g linoleic acid $\left({\text{C}}_{18}{\text{H}}_{32}{\text{O}}_2\right)$ to stearic acid $\left({\text{C}}_{18}{\text{H}}_{36}{\text{O}}_2\right)$.

Short answer

The volume of hydrogen required is 8.093 L.

Step-by-step solution

Balanced chemical equation

When linoleic acid reacts with hydrogen, it produces stearic acid.

The balanced chemical equation of the conversion of linoleic acid to stearic acid is given below:

${\text{C}}_{18}{\text{H}}_{32}{\text{O}}_2+2{\text{H}}_2\to {\text{C}}_{18}{\text{H}}_{36}{\text{O}}_2$

Step 2: Calculation of number of moles of linoleic acid

Given that the mass of linoleic acid,

${\text{m}}_{\text{lino}}=500.0\text{g}$

Molar mass of linoleic acid,

${\text{M}}_{\text{lino}}\text{= 280.4472}{\text{gmol}}^{-1}$

Pressure,

$\text{P = 1.00}\text{atm}$

Temperature,

$\text{T =0}°\text{C = 273}\text{K}$

Number of moles of linoleic acid,

$\begin{array}{rcl}{\mathrm{n}}_{\mathrm{lino}}& =& \frac{{\mathrm{m}}_{\mathrm{lino}}}{{\mathrm{M}}_{\mathrm{lino}}}\\ & =& \frac{500.0}{280.4472}=1.7829\mathrm{mol}\\ & & \end{array}$

Calculation of volume of hydrogen

1 mole of linoleic acid required two moles of hydrogen to convert to stearic acid. So 1.7829 moles of linoleic acid requires, $\left(2\times 1.7829\right)\text{mol}$ of hydrogen gas, i.e., $3.5658\text{mol}$ of hydrogen gas.

Number of moles of hydrogen,

${\text{n}}_{\text{H}}=3.5658\text{mol}$

Now, let us use ideal gas equation to find the volume of hydrogen gas.

$\text{PV = nRT}$

Substituting the values, we get

$\begin{array}{rcl}1\mathrm{atm}\times \mathrm{V}& =& 3.5658\mathrm{mol}\times 8.314{\mathrm{Jmol}}^{-1}{\mathrm{K}}^{-1}\times 273\mathrm{K}\\ \mathrm{V}& =& \frac{3.5658\mathrm{mol}\times 8.314{\mathrm{Jmol}}^{-1}{\mathrm{K}}^{-1}\times 273\mathrm{K}}{1\mathrm{atm}}\\ & =& 8.093\mathrm{L}\end{array}$

Thus, the volume of hydrogen is 8.093 L.