Question

Methanol CH3OH is the simplest of the alcohols . It is sinthesized by the reaction of hydrogen and carbon monoxide.

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

a) If 500 ml CO and 750 mol H2 are present, which is the limiting reactant?

b) How many mols of excess reactant remain unchange?

c) How many moles of CH3OH are formed ?

Answer 1

a) CO is the limiting reactant.

b)
`n_(H_2)^(excess)=749.9592molH_2`

c)
`n_(CH_3OH)=0.0204molCH_3OH`

Step-by-step explanation:

Hello,

In this case, the balanced chemical reaction:

`CO + 2H_2 \rightarrow CH_3OH`

So we proceed as follows:

a) We first compute the moles of CO by using its density (0.00114 g/mL) and molar mass (28 g/mol):

`n_(CO)=500mLCO*(0.00114gCO)/(1mLCO)*(1molCO)/(28gCO)=0.0204molCO`

Next, with the given 750 moles of hydrogen, we can compute the moles of carbon monoxide that are consumed by such amount of hydrogen by using their 1:2 molar ratio:

`n_(CO)^(consumed\ by\ H_2)=750molH_2*(1molCO)/(2molH_2)=375molCO`

Thus, we see a clear excess of hydrogen, for that reason the carbon monoxide is the limiting reactant.

b) In this case, we first compute the moles of CO that are not consumed:

`n_(CO)^(unchanged)=375mol-0.0204mol=374.9796molCO`

Next, we use the 1:2 molar ratio again to compute the unchanged moles of hydrogen which is the excess reactant:

`n_(H_2)^(excess)=375.9796molCO*(2molH_2)/(1molCO) =749.9592molH_2`

c) Finally, we use the reacting moles of carbon monoxide to compute the formed moles of methanol by using the 1:1 molar ratio:

`n_(CH_3OH)=0.0204molCO*(1molCH_3OH)/(1molCO) =0.0204molCH_3OH`

Best regards.