Question

Methanol is the simplest alcohol and it is used as a fuel in race cars and is a potential replacement for gasoline. Methanol can be manufactured by combination of gaseous carbon monoxide and hydrogen – if 68.5 Kg of carbon monoxide are reacted with 8.6 Kg of hydrogen, calculate the theoretical yield of methanol. If 3.57* 104 g of methanol are actually produced, what is the percent yield of methanol?

Answer 1

balanced equation :

CO       +     2H2   -------------------> CH3OH

28g          4g                                  32.0g

68500 g      8600 g

28 g CO ----------------> 4 g H2

68500 g CO ---------------> 4 x 68500 / 28 = 9786 g H2 needed

but we have only 8600 g H2. so limiting reagent is H2

mass of CH3OH = 8600 x 32 / 4

                          = 68800 g

theoretical mass = 68800 g

actal mass = 3.57 x 104 g

percent yield of methanol = actual mass x 100 / theoretical mass

                                          = 3.57 x 104 x 100 / 68800

                                         = 51.9 %