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

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.