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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.
Answer:
Explanation:
Our sample yielded 1g of
and 16g of
, but our unknown sample yielded 2 times as much
for the same amount of
.
What does this mean? that the H:O proportion for the unknown sample is twice the H:O proportion for the sample.
What is the H:O proportion for the sample? As we can see from its formula, it's 1:1, therefore the proportion for the unknown formula must be 2:1.
That means, two H atoms for every O atom. We could write that as: and you should recognize that formula, for it is one of the most common compounds on earth, Water.