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17) 4NH3 + 502 → 4NO + 6H2OIf 73 grams of NH3 are reacted and 101 grams of H20 are actually produced, what is the percentyield?

User Topper
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Answer

%Yield = 87.2%

Step-by-step explanation

Given:

mass of NH3 reacted = 73 g

mass of H2O produced = 101 g

We know:

molar mass of NH3 = 17,031 g/mol

molar mass of water = 18.01528 g/mol

Required: % Yield

Solution:

The formula used to calculate the percentage yield is:

%Yield = (actual yield/theoretical yield) x 100

The actual yield of H2O produced = 101 g

Now lets calculate the theoretical yield first.

First find the number of moles of NH3, and use stoichiometry to find the theoretical mass of water.

n = m/M n is the moles, m is the mass and M is the molar mass.

n = 73/17,031 g/mol

n = 4.29 mol

Using the stoichiometry, there molar ratio between NH3 and H2O is 4:6

Therefore the moles of H2O = 4.29 x (6/4)

n of H2O = 6.43 mol

The theoretical mass can then be calculated:

m = n x M

m = 6.43 mol x 18.01528 g/mol

m = 115.83 g

%Yield = (101 g/115.83)*100

%Yield = 87.2%

User Egilhh
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