Answer:
The answer to your question is Molarity = 0.41
Explanation:
Data
mass of KNO₃ = 76.6 g
volume = 1.84 l
density = 1.05 g/ml
Process
1.- Calculate the molecular mass of KNO₃
molecular mass = 39 + 14 + (16 x 3) = 101 g
2.- Calculate the number of moles
101 g of KNO₃ --------------- 1 mol
76.6 g of KNO₃ ------------ x
x = (76.6 x 1) / 101
x = 0.76 moles
3.- Calculate molarity
Molarity =
Substitution
Molarity =
Result
Molarity = 0.41
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Answer:
36.37% is the percent yield of the reaction.
Explanation:
1)0.650 L nitrogen gas , at 295 K and 1.01 bar.
Let the moles of nitrogen gas be n.
Pressure of the gas ,P= 1.01 bar = 0.9967 atm (1 bar = 0.9869 atm)
Temperature of the gas = T = 295 K
Volume of the gas = V = 0.650 L
Using an ideal gas equation:
2) Moles of ammonia gas=
Moles of oxygen gas =
According to reaction ,3 mol of oxygen reacts with 4 mol of ammonia.
Then,0.1101 mol of oxygen will react with:
of ammonia.
Hence, oxygen gas is in limiting amount and act as limiting reagent.
3) Theoretical yield of nitrogen gas :
According to reaction, 3 mol of oxygen gas gives 2 moles of nitrogen gas.
Then 0.1101 mol of oxygen will give:
of nitrogen.
Theoretical yield of nitrogen gas = 0.0734 mol
Experimental yield of nitrogen as calculated in part (1) = 0.0267 mol
Percentage yield:
Percentage yield of the reaction:
36.37% is the percent yield of the reaction.