Answer : The limiting reagent is 
Solution : Given,
Moles of methane = 2.8 moles
Moles of = 5 moles
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 mole of react with 1 mole of 
So, 5 moles of react with 
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.
Hence, the limiting reagent is
Answer:
Don't mark me brainliest because of this but I'm pretty sure your supposed to give us the words because teachers don't give you things like that without the words you will the answer in with.
Answer:
P' = 41.4 mmHg → Vapor pressure of solution
Explanation:
ΔP = P° . Xm
ΔP = Vapor pressure of pure solvent (P°) - Vapor pressure of solution (P')
Xm = Mole fraction for solute (Moles of solvent /Total moles)
Firstly we determine the mole fraction of solute.
Moles of solute → Mass . 1 mol / molar mass
20.2 g . 1 mol / 342 g = 0.0590 mol
Moles of solvent → Mass . 1mol / molar mass
60.5 g . 1 mol/ 18 g = 3.36 mol
Total moles = 3.36 mol + 0.0590 mol = 3.419 moles
Xm = 0.0590 mol / 3.419 moles → 0.0172
Let's replace the data in the formula
42.2 mmHg - P' = 42.2 mmHg . 0.0172
P' = - (42.2 mmHg . 0.0172 - 42.2 mmHg)
P' = 41.4 mmHg
