Question

Methanol, CH₃OH, can be made from the reaction of carbon monoxide and hydrogen.

CO(g) + 2 H₂(g) → CH₃OH(l)

What mass of hydrogen is required to produce 3.50 L of methanol (d = 0.791 g/mL) if this reaction has a 73.01% yield under certain conditions?

Answer 1

Solution :-

3.50 L * 1000 ml / 1 L = 3500 ml methanol

Mass of methanol = volume * density

                                 = 3500 ml * 0.791 g per ml

                                = 2768.5 g

Now lets calculate the moles of methanol

Moles of methanol = 2768.5 g / 32.04 g per mol

                                   = 86.41 mol methanol

Lets calculate the thereotical moles because the yield is only 73.01 %

86.41 mol * 100 % / 73.01 % = 118.35 mol methanol

Now using the moles ratio lets calculate the moles of the H2 needed for the reaction

118.35 mol methanol * 2 mol H2 / 1 mol methanol = 236.7 mol H2

Now lets convert moles of H2 to its mass

Mass of H2 = moles * molar mass

                     = 236.7 mol H2 * 2.01588 g per mol

                     = 477 g H2

So the mass of the H2 needed = 477 g H2