23.9 11.00 would be the wrong answer because I guessed
Methanol is prepared by reacting Carbon monoxide and Hydrogen gas,
CO + 2 H₂ → CH₃OH
Calculating Moles of CO:
According to equation,
32 g (1 mole) of CH₃OH is produced by = 1 Mole of CO
So,
3.60 × 10² g of CH₃OH is produced by = X Moles of CO
Solving for X,
X = (3.60 × 10² g × 1 Mole) ÷ 32 g
X = 11.25 Moles of CO
Calculating Moles of H₂:
According to equation,
32 g (1 mole) of CH₃OH is produced by = 2 Mole of H₂
So,
3.60 × 10² g of CH₃OH is produced by = X Moles of H₂
Solving for X,
X = (3.60 × 10² g × 2 Mole) ÷ 32 g
X = 22.5 Moles of H₂
Result:
3.60 × 10² g of CH₃OH is produced by reacting 11.25 Moles of CO and 22.5 Moles of H₂.
I think is 1 and a half km
