Respuesta :

We are given with the reaction that produces methanol from the reaction of hydrogen gas and carbon monoxide. This is expressed in the balanced equation: 2H2+CO=CH3OH. We need to identify the limiting reactant. Convert each mass to moles and divide each with their corresponding stoich. coeff. For H2, this is equal to 3 and for CO, this is equal to 2.66. Hence CO is the limiting reactant. From this, the amount of methanol produced is 85.14 grams. 

Answer : The theoretical yield of the methanol is, 85.12 g

Solution : Given,

Mass of [tex]H_2[/tex] = 12 g

Mass of [tex]CO[/tex]= 74.5 g

Molar mass of [tex]H_2[/tex] = 2 g/mole

Molar mass of [tex]CO[/tex] = 28 g/mole

Molar mass of [tex]CH_3OH[/tex] = 32 g/mole

First we have to calculate the moles of [tex]H_2[/tex] and [tex]CO[/tex].

Moles of [tex]H_2[/tex] = [tex]\frac{\text{ given mass of }H_2}{\text{ molar mass of }H_2}= \frac{12g}{2g/mole}=6moles[/tex]

Moles of [tex]CO[/tex] = [tex]\frac{\text{ given mass of }CO}{\text{ molar mass of }CO}= \frac{74.5g}{28g/mole}=2.66moles[/tex]

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be,

[tex]CO+2H_2\rightarrow CH_3OH[/tex]

From the balanced reaction we conclude that

As, 1 mole of CO react with 2 moles of [tex]H_2[/tex]

So, 2.66 moles of CO react with [tex]2\times 2.66=5.32moles[/tex] of [tex]H_2[/tex]

Excess moles of [tex]H_2[/tex] = 6 - 5.32 = 0.68 moles

That means CO is a limiting reagent and [tex]H_2[/tex] is an excess reagent.

Now we have to calculate the moles of methanol.

From the reaction we conclude that,

As, 1 mole of CO react to give 1 mole of methanol

So, 2.66 moles of CO react to give 2.66 moles of methanol

Now we have to calculate the mass of methanol.

[tex]\text{Mass of }CH_3OH=\text{Moles of }CH_3OH\times \text{Molar mass of }CH_3OH[/tex]

[tex]\text{Mass of }CH_3OH=(2.66moles)\times (32g/mole)=85.12g[/tex]

Therefore, the theoretical yield of the methanol is, 85.12 g