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.
2.5X20=50g
50g should be the right answer
Mass=volumeXdenisty.
Answer: Rate law is, R= 6.02[A]¹[B]²
Explanation in attached image
Answer is: osmotic pressure.
Osmotic pressure, alongside the vapor pressure depression, freezing point depression and the boiling point elevation are<span> the </span>colligative properties od solution.
<span>The direction of osmotic pressure is always from the side with the lower concentration (c = n/V) of solute to the side with the higher concentration.</span>
It requires a force in the direction opposite to the motion of the object for it to slow down.
