If the reaction of 77.0 grams of CaCN2 produces 27.1 grams of NH3, what is the percent yield? CaCN2+3H2O=>CaCO3+2NH3
Answers
Answer:
82.74%
Explanation:
Molar mass of CaCN₂ = 40 + 12 + 2 x 14 = 80 g/mol
Molar mass of NH₃ = 14 + 3 x 1 = 17 g/mol
The given reaction is
CaCN₂ + 3 H₂O ⇒ CaCO₃ + 2 NH₃
From the reaction, we can conclude that:
1 mole of CaCN₂ produces 2 mole of NH₃
In terms of mass,
⇒80 g of CaCN₂ produces 34 g of NH₃
⇒77 g of CaCN₂ produces (34/80) x 77 = 32.725 g of NH₃
⇒Theoretical yield of the reaction = 32.725 g
According to question, mass of NH₃ obtained = 27.1 g
Hence, Percent yield = (Actual mass obtained/Theoretical yield) x 100
= (27.1/32.725) x 100
= 82.74%
Answer: The answer is 83.3 %.
Explanation:
From the balanced chemical equation
CaCN2 +
3H2O
CaCO3
+
2NH3
we can see that we have a 1: 2 mole ratio between CaCN2 and NH3; that is, for every mole of CaCN2 used in the reaction, 2 moles of NH3 will be formed.
In order to determine the percent yield, we must determine what the limiting reagent is. We know that CaCN2 and NH3's molar masses are
80 and 17.
mol
nCaCN
moles and
nNH3
mol
mCaCN2
respectively, therefore
779
80.0 9 mol
27.1g
17 mol
= 1.6 moles
molarmass
= 0.96
MNH3
molarmass
According to our mole-to-mole ratio, 0.96 moles of CaCN2 should have produced 20.96 1.92 moles of NH3, but instead
produced 1.6 moles. This means that not all CaCN2 reacted <=> water is the limiting
reagent.
So, the percent yield, defined as the actual yield
divided by the theoretical yield and multiplied by 100%, is
% yield
- 100% =
actual
theoretic
83.3%
100%
1.6moles 1.92moles