Answer:
Explanation:
Here we have to use stoichiometry.
First of all, we have to calculate the mass of 100% of yield:
1.7 g ------- 98%
X -------- 100%
X = 1.73 g (approximately)
Second, we have to calculate the mass of N2 that is necessary to react to produce the mass of 1.73g of NH3. To do that, we have to use the Molar mass of N2 and NH3 and don't forget the stoichiometric relationship between them.
Molar Mass N2 : 14x2 = 28 g/mol
Molar Mass NH3: 14 + 3 = 17 g/mol
28g (N2) ------- 17x2 (NH3)
X ------------ 1.73 g
X = 1.42 g (approximately)
Molarity: M = #moles of solute / liters of solution
# moles = mass / molar mass
Molar mass calculation
Barium hydroxide = Ba (OH)2
Atomic masses
Ba = 137.4 g/mol
O=16 g/mol
H=1 g/mol
Molar mass of Ba (OH)2 = 137.4 g/mol + 2*16g/mol + 2*1 g/mol = 171.4 g/mol
# mol = 25.0g/171.4 g/mol = 0.146 mol
For the volume of water use the fact that the density is 1g/ml., so 120 g = 120 ml = 0,120 liters.
M = 0.146mol / 0.120 liters = 1.22 mol/liter
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