Methanol, ch3oh, is a useful fuel that can be made as follows: co(g) + 2h2(g) → ch3oh(l) a reaction mixture used 12.0 g of h2 and 74.5 g of co. (a) determine the theoretical yield of ch3oh. (b) calculate the amount of the excess reactant that remains unchanged at the end of the reaction.

The theoretical yield of methanol is 85 g and 1.36 g of excess reactant is left over.

The equation of the reaction is; CO(g) + 2H2(g) → CH3OH(l)

Number of moles of H2 =  12.0 g/2.00 g/mol = 6 moles

Number of moles of CO =  74.5 g /28 g/mol = 2.66 moles

1 mole of CO reacts with 2 moles of H2

2.66 moles of CO reacts with 2.66 moles ×  2 moles/1 mole

= 5.32 moles of H2

H2 is the reactant in excess because there is more than enough H2 to react with CO.

Hence;

1 mole of CO produces 1 mole of methanol

2.66 moles of CO produces 2.66 moles of methanol

Theoretical yield of methanol = 2.66 moles of methanol × 32 g/mol = 85 g

The excess reactant is H2. Amount of excess H2 is obtained from;

Amount of H2 present - amount of H2 reacted

6 .00 moles - 5.32 moles  = 0.68 moles

Mass of unreacted H2 = 0.68 moles × 2.00 g/mol = 1.36 g of H2

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A = Theoretical yield = 82.24 g

B = Amount of excess reactant left = 1.72 g

Explanation:

Given data:

Mass of H₂ = 12 g

Mass of CO = 74.5 g

Theoretical yield of CH₃OH = ?

Amount of excess reactant left = ?

Solution:

First of all we will write the balance chemical equation:

CO + 2H₂  →   CH₃OH

Number of moles of H₂ = mass / molar mass

Number of moles of H₂ = 12 g/ 2 g/mol = 6 mol

Number of moles of CO = mass / molar mass

Number of moles of CO = 74.5 g/ 29 g/mol = 2.57 mol

Now we compare the moles of CH₃OH  with CO and H₂ from balance chemical equation:

                                              CO         :      CH₃OH  

                                                 1           :         1

                                                 2.57      :        2.57

                                                  H₂        :      CH₃OH

                                                   2          :            1

                                                    6         :          1/2 × 6 = 3 mol

The number of moles of CH₃OH produce by  CO are less so it will limiting reactant.

mass of CH₃OH  = number of moles × molar mass

mass of CH₃OH  =   2.57 mol ×   32 g/mol

mass of CH₃OH  =    82.24 g

Excess amount of H₂:

                                     CO         :         H₂

                                      1            :            2

                                  2.57         :             2×2.57 = 5.14 mol

The moles of H₂ that react with CO are 5.14. While the total number of moles of H₂ available are 6 moles. So,

The number of moles of H₂ remain untreated = 6 mol - 5.14 mol = 0.86 mol

Mass of H₂ remain untreated =  0.86 mol × 2 g/mol

Mass of H₂ remain untreated =  1.72 g

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