Answer:
1. 80.9 g of CH₃OH
2. CO
3. 5.14moles
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
