Question

In the equation: 4NH3 + 5O2 → 4NO + 6H2O

, 310 g of O2 will react with 175 g of NH3. What is the limiting reactant? What is the theoretical yield of NO, and if 197 g of NO are produced, what is the percent yield?

Answer 1

Theoretical yield = 202g; Percentage yield = 98%

Explanation:

`Mr NH_3 = 17; Mr O_2 = 32; Mr NO = 26; Mr H_2O = 18`

Using the equation:

Number of Moles = Mass / Mr

Number of Moles `NH_3 = 175/17 = 10.3`
Number of Moles `O_2 = 310/32 = 9.69`

`O_2` is the limiting reagent because the least moles of this are used in the reaction.

From the balanced equation, we can see that `O_2` and `NO` are in a 5:4 ratio

`5/4 = 0.8`

`9.69 xx 0.8 = 7.75`

7.75 moles of `NO` is the maximum that can be produced from 310g of `O_2`. To convert this to mass we use the same equation as in the first step.

Mass = Moles x Mr

`7.75 xx 26 = 202g`

202g is the theoretical yield of `NO`

To calculate the percentage yield, you just divide the actual yield by the theoretical yield and multiply by 100:
`197 / 202 xx 100 = 98%`