Question

What is the percent yield when 100.0 g of Mg3N2 reacts with 75.0 g H2O to produce 15.0 g ofNH3? Show all your workMg3N2(s) + 6H2O(1) ► 3Mg(OH)2(aq) + 2NH3(8)

Answer 1

Answer:

The percent yield is 48.29%.

Explanation:

1st) To calculate the percent yield, it is necessary to use the Actual yield (the given values in the exercise) and the Theoretical yield (the values given by the balanced chemical reaction).

Balanced chemical reaction:

$Mg_3N_2+6H_2O\rightarrow3Mg(OH)_2+2NH_3$

With the balanced chemical reaction we know that 1 mole of Mg3N2 reacts with 6 moles of H2O to produce 3 moles of Mg(OH)2 and 2 moles of NH3.

2nd) It is necessary to convert the moles to grams, using the molar mass of each compound:

- Mg3N2 molar mass: 100.95g/mol

- H2O molar mass: 18g/mol

- NH3 molar mass: 17.03g/mol

- Conversion of Mg3N2 moles to grams:

$1mol*\frac{100.95g}{1mol}=100.95g$

- Conversion of H2O moles to grams:

$6moles*\frac{18g}{1mol}=108g$

- Conversion of NH3 moles to grams:

$2moles*\frac{17.03g}{1mol}=34.06g$

Now we know that 100.95g of Mg3N2 react with 108g of H2O to produce 31.06g of NH3. Those values are the Theorerical values.

3rd) To calculate the percent yield we can use the formula and replace the values of the product NH3:

$\begin{gathered} \text{ Percent yield=}\frac{\text{ Actual yield}}{\text{ Theoretical yield}}*100\% \\ \text{ Percent yield }=\text{ }\frac{\text{ 15.0g }}{\text{ 31.06g}}*100\% \\ \text{ Percent yield=48.29}\% \\ \end{gathered}$

So, the percent yield is 48.29%.