Question

If the reaction of 77.0g of CaCN2 produces 27.1g of NH3, what is the percent yield?

Answer 1

Answer:

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Explanation:

The answer is 83.3%

Answer 2

Explanation:

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%