Answer:
53.5g of NH4Cl
Explanation:
First, we need to obtain the number of mole of NH4Cl. This is illustrated below:
Volume = 0.5L
Molarity = 2M
Mole =?
Molarity = mole /Volume
Mole = Molarity x Volume
Mole = 2 x 0.5
Mole = 1mole
Now, let us convert 1mole of NH4Cl to gram. This is illustrated below:
Molar Mass of NH4Cl = 53.5g/mol
Number of mole = 1
Mass =?
Number of mole = Mass /Molar Mass
Mass = number of mole x molar Mass
Mass = 1 x 53.5
Mass = 53.5g
Therefore, 53.5g of NH4Cl is contained in the solution.
The balanced equation for the formation of ammonia is as follows
N₂ + 3H₂ ---> 2NH₃
stoichiometry of N₂ to H₂ is 1:3
we need to find the moles of N₂, volume of N₂ has been given
molar volume is where 1 mol of any gas occupies a volume of 22.4 L at STP.
if 22.4 L is occupied by 1 mol
then 3.5 L of gas is occupied by - 3.5 L / 22.4 L/mol = 0.16 mol
number of moles of N₂ present - 0.16 mol
1 mol of N₂ requires 3 mol of H₂
therefore 0.16 mol of N₂ requires - 3 x 0.16 = 0.48 mol of H₂
mass of H₂ required - 0.48 mol x 2 g/mol = 0.96 g
0.96 g of H₂ is required
Natural abundance of oxygen I think
