# 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?

Jan 9, 2018

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 $N {H}_{3} = \frac{175}{17} = 10.3$
Number of Moles ${O}_{2} = \frac{310}{32} = 9.69$

${O}_{2}$ 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 $N O$ are in a 5:4 ratio

$\frac{5}{4} = 0.8$

$9.69 \times 0.8 = 7.75$

7.75 moles of $N O$ 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

$7.75 \times 26 = 202 g$

202g is the theoretical yield of $N O$

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