Question

Nitric oxide, no, is made from the oxidation of nh3, and the reaction is represented by the equation 4nh3 + 5o2 ? 4no + 6h2o an 9.5-g sample of nh3 gives 12.0 g of no. the percent yield of no is :

Answer 1

4NH₃ + 5O₂ = 4NO + 6H₂O

m(NH₃)=9.5 g
m'(NO)=12.0 g

the theoretical mass:
m(NO)=M(NO)m(NH₃)/M(NH₃)

w=m'(NO)/m(NO)

w=m'(NO)M(NH₃)/{M(NO)m(NH₃)}

w=12.0*17.03/{30.01*9.5}=0.7168 (71.68%)

Answer 2

Answer: The percent yield of NO is 71.43 %.

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

• For ammonia:

Given mass of ammonia = 9.5 g

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

`\text{Moles of ammonia}=(9.5g)/(17g/mol)=0.56mol`

The given chemical equation follows:

`4NH_3+5O_2\rightarrow 4NO+6H_2O`

By Stoichiometry of the reaction:

4 moles of ammonia produces 4 moles of NO

So, 0.56 moles of ammonia will produce
`(4)/(4)* 0.56=0.56mol` of NO

Now, calculating the mass of NO from equation 1, we get:

Molar mass of NO = 30 g/mol

Moles of NO = 0.56 moles

Putting values in equation 1, we get:

`0.56mol=\frac{\text{Mass of NO}}{30g/mol}\\\\\text{Mass of NO}=(0.56mol* 30g/mol)=16.8g`

• To calculate the percentage yield of NO, we use the equation:

`\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}* 100`

Experimental yield of NO = 12.0 g

Theoretical yield of NO = 16.8 g

Putting values in above equation, we get:

`\%\text{ yield of NO}=(12.0g)/(16.8g)* 100\\\\\% \text{yield of NO}=71.43\%`

Hence, the percent yield of NO is 71.43 %.