Question

1. For the following reaction, 4.86 g of magnesium nitride are mixed with excess water. The reaction yields 7.18 g of magnesium hydroxide.

magnesium nitride(s) + water(1) –> magnesium hydroxide (aq) + ammonia (aq)
What is the ideal yield of magnesium hydroxide?
What is the percent yield for this reaction?
2. For the following reaction, 6.41 g of hydrogen gas are mixed with excess nitrogen gas. The reaction yields 26.2 g of ammonia.
nitrogen(g) + hydrogen(g) –> ammonia(g)
What is the ideal yield of ammonia?
What is the percent yield for this reaction?
3. For the following reaction, 3.79 g of water are mixed with excess chlorine gas. The reaction yields 8.70 g of hydrochloric acid.
chlorine(g) + water(1) –> hydrochloric acid(aq) + chloric acid (HCIO3)(aq)
What is the ideal yield of hydrochloric acid?
What is the percent yield for this reaction?

Answer 1

See explanation

Step-by-step explanation:

1)

Mg3N2(s) + 6H2O(l) ------------> 3Mg(OH)2 + 2NH3(g)

Number of moles of magnesium nitride= mass/molar mass= 4.86g/100.9494 g/mol = 0.048 moles

1 mole of magnesium nitride yields 3 moles of magnesium hydroxide

0.048 moles of magnesium nitride yields 0.048 moles × 3= 0.144 moles of magnesium hydroxide

Theoretical yield of magnesium hydroxide = 0.144 moles × 58.3197 g/mol = 8.398 g

Percent yield= actual yield/ theoretical yield × 100

Percent yield= 7.18/8.398 × 100/1 = 85.5%

2)

N2(g) + 3H2(g) -------> 2NH3(g)

Number of moles of hydrogen gas = mass/ molar mass = 6.41g/ 2gmol-1 = 3.205 moles of hydrogen gas.

From the balanced reaction equation;

3 moles of hydrogen gas yields 2 moles of ammonia

3.205 moles of hydrogen gas yields 3.205 × 2/3 = 2.1367 moles of ammonia

Theoretical yield of ammonia = 2.1367 moles × 17 gmol-1 = 36.3 g

Percent yield = actual yield/ theoretical yield ×100

Percent yield = 26.2/36.3 ×100

Percent yield = 72.2%

3)

3Cl2(g) + 3H2O(l) ------> HOCl3(aq) + 5HCl(aq)

Number of moles of water= mass/ molar mass = 3.79g/18 gmol-1 = 0.21 moles

Since

3 mole of water yields 5 mole of HCl

0.21 moles of water yields 0.21 × 5/3 = 0.35 moles of HCl

Theoretical yield of HCl = 0.35 moles × 36.5 gmol-1 = 12.775 g

Percent yield = actual yield/ theoretical yield × 100/1

Percent yield = 8.70/12.775 ×100

Percent yield = 68.1%