Methanol, ch, oh (mm = 32.05 g/mol), is prepared industrially from this gas-phase reaction:
co(g) + 2h2(g) - ch2oh()
in a laboratory test, a reaction vessel was filled with 63.4 g co (mm = 28.01 g/mol) and 97.2 g of hz.
how many grams of methanol would be produced in a complete reaction? which reactant remains left
over at the end of the reaction? how many grams of it remain?

Hydrogen will left.

Mass of hydrogen left = 88 g            

Mass of methane produced = 73.72 g

Explanation:

Given data:

Molar mass of methanol = 32.05 g/mol

Mass of carbon monoxide = 63.4 g

Molar mass of carbon monoxide = 28.01 g/mol

Mass of hydrogen gas = 97.2 g

Molar mass of hydrogen gas = 2 g/mol

Mass of  methanol = ?

Excess reactant = ?

Amount of excess reactant left = ?

Solution:

Chemical equation:

CO(g) + 2H₂(g)   → CH₃OH(g)

Number of moles of CO

Number of moles = mass/ molar mass

Number of moles = 63.4 g/ 28.01 g/mol

Number of moles = 2.3 mol

Number of moles of H₂:

Number of moles = mass/ molar mass

Number of moles = 97.2 g/ 2 g/mol

Number of moles = 48.6  mol

Now we compare the moles of methane with hydrogen and carbon monoxide gas through balanced chemical equation.

                       CO     :     CH₃OH

                        1        :          1

                        2.3    :        2.3

                      H₂        :        CH₃OH

                        2        :            1

                       48.6    :            1/2×48.6 = 24.45 mol

Number of moles of methane produced by CO are less so it will limiting reactant.

Mass of methane = moles × molar mass

Mass of methane = 2.3 mol × 32.05 g/mol

Mass of methane = 73.72 g

The hydrogen gas is in excess amount. Some of its amount will remain at the end of reaction unreacted.

Amount of hydrogen remain unreacted:

                              CO          :        H₂

                                1            :         2

                               2.3         :         2×2.3 = 4.6 mole

so 4.6 moles of H₂ react with 2.3 moles of CO.

Total number of moles of hydrogen = 48.6 mole

Number of moles left = 48.6 - 4.6 = 44 moles

Mass of hydrogen left = moles  × molar mass

Mass of hydrogen left = 44 mol  × 2g/mol

Mass of hydrogen left = 88 g