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}= (12g)/(2g/mole)=6moles`

Moles of
`CO` =
`\frac{\text{ given mass of }CO}{\text{ molar mass of }CO}= (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* 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* \text{Molar mass of }CH_3OH`

`\text{Mass of }CH_3OH=(2.66moles)* (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.