Question

70.8 g of CO is exposed 20.4 g of Hy and a reaction takes place.

CO + 2H2 - CH3OH
1. How many grams of methyl hydroxide (CH3OH) is produced?
2. What is the limiting reactant?
3. How much of the excess reactant is left unused?

literally what does any of this mean​

Answer 1

1. 80.9 g of CH₃OH

2. CO

3. 5.14moles

Step-by-step explanation:

The reaction is:

CO + 2H₂ → CH₃OH

First place, we determine the moles of each reactant.

70.8 g . 1mol/ 28g = 2.53 moles of CO

20.4 g . 1mol / 2g = 10.2 moles of H₂

There is too much hydrogen, so the limiting reactant might be the carbon monoxide.

2 moles of H₂ react to 1 mol of CO

Then, 10.2 moles of H₂ may react to (10.2 . 1) /2 = 5.1 moles

We don't have enough CO, so it's ok that CO is the limiting. (We have 2.53 moles, we need 5.1 moles)

In the other hand, hydrogen is the excess reactant.

1 mol of CO react to 2 moles of hydrogen

2.53 moles may react to (2.53 . 2) /1 =5.06 moles

We have 10.2 moles, and we need 5.06 moles. Then (10.2 - 5.06) = 5.14 moles remains after the reaction goes complete.

Now that we know the limiting reactant, we can determine the grams of produced methanol. Ratio is 1:1

1 mol of CO produces 1 mol of CH₃OH

2.53 moles of CO must produce 2.53 moles of CH₃OH

We convert moles to mass: 2.53 mol . 32 g/mol = 80.9 g