Answer:
The answer to your question is: 2 grams of methane
Explanation:
Data
V = 9.15 l
P = 1.77 atm
T = 57° C = 330 °K
Ar mass = 19 g
CH4 mass = ?
Formula
PV = nRT
Process
n = 0.6
Argon
40 g of Ar -------------------- 1 mol
19 g --------------------- x
x = 0.475 mol of Ar
moles of CH4 = 0.6 - 0.475
= 0.125
Methane
16 g of CH4 --------------- 1mol
x ---------------- 0.125 mol
x = 2 grams
Answer:
moles of ammonia produced = 0.28 moles
Explanation:
The reaction is
As per equation, one mole of nitrogen will react with three moles of hydrogen to give two moles of ammonia
So 0.140 moles of nitrogen will react with = 3 X 0.140 moles of Hydrogen
= 0.42 moles of hydrogen molecule.
this will give 2 X 0.140 moles of ammonia = 0.28 moles of ammonia
the moles of ammonia produced = 0.28 moles
Here the nitrogen is limiting reagent.
Iron has magnetic properties, so you would just need to hold a magnet over the mixture and the filings will come right out.
