So if we use the equation:
→ 
We can then determine the amount of 
needed to produce 208 kg of methanol.
So let's find out how many moles of methanol 208 kg is:
Methanol molar weight = 32.041g/mol
So then we can solve for moles of methanol:
So now that we have the amount of moles produced, we can use the molar ratio (from the balanced equation) of hydrogen and methanol. This ratio is 2:1 hydrogen:methanol.
Therefore, we can set up a proportion to solve for the moles of hydrogen needed:
So now that we have the number of moles of 
that are produced, we can then use the molar weight of hydrogen to solve for the mass that is needed:
Therefore, the amount of diatomic hydrogen () that is needed to produce 208kg of methanol is 
g.
Answer:
Vapor pressure of Eugenol = 667.04torr
Explanation:
By applying dalton's law of partial pressure;
at 100degree celsius, total pressure = 744.67torr
vapor pressure of water at 100 degree celsius = 760torr
mole fraction of eugenol = 0.1648
mole fraction of water = 1 - 0.1648 = 0.8352
Total pressure = vapor P(water) + vapor P( Eugenol)
for water; vapour pressure = mole fraction x total pressure
for eugenol; vapor pressure = mole fraction x total pressure
substituting into the above gives the vapor pressure of eugenol = 667.04torr
High values of Ksp greater than 1 infer that the solid phase is relatively soluble and the chemical reaction proceeds readily to the right. This is because the formula for ksp is a ratio between the products and reactants. A ksp value greater than 1 means that there are more products than reactants; thus, favoring the forward reaction.
It’s Tetracarbon dioxide :)
