Question

What is the theoretical yield of methanol (ch3oh) when 12.0 grams of h2 is mixed with 74.5 grams of co? co 2h2 ch3oh

Answer 1

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

Solution : Given,

Mass of $H_2$ = 12 g

Mass of $CO$= 74.5 g

Molar mass of $H_2$ = 2 g/mole

Molar mass of $CO$ = 28 g/mole

Molar mass of $CH_3OH$ = 32 g/mole

First we have to calculate the moles of $H_2$ and $CO$.

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

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

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be,

$CO+2H_2\rightarrow CH_3OH$

From the balanced reaction we conclude that

As, 1 mole of CO react with 2 moles of $H_2$

So, 2.66 moles of CO react with $2\times 2.66=5.32moles$ of $H_2$

Excess moles of $H_2$ = 6 - 5.32 = 0.68 moles

That means CO is a limiting reagent and $H_2$ 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.

$\text{Mass of }CH_3OH=\text{Moles of }CH_3OH\times \text{Molar mass of }CH_3OH$

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

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

Answer 2

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.