Answer to Question #267849 in General Chemistry for Roseanne

Question

Aqueous sulfuric acid H

2

SO

4

 reacts with solid sodium hydroxide NaOH

 to produce aqueous sodium sulfate Na

2

SO

4

 and liquid water H

2

O

. If 10.6

g

 of sodium sulfate is produced from the reaction of 8.8

g

 of sulfuric acid and 9.1

g

 of sodium hydroxide, calculate the percent yield of sodium sulfate. 

Round your answer to 2

 significant figures. 

Answer 1

The equation is below:

H₂SO₄(aq) + 2NaOH(s) --> Na₂SO₄(aq) + 2H₂O(l).

The limiting reactant should be determined first.

$n(H_2SO_4)=8.8\ g\times\frac{1\ mol}{98.08\ g}=0.08972\ mol$

$n(NaOH)=9.1\ g\times\frac{1\ mol}{40.00\ g}=0.2275\ mol$

According to the equation, NaOH and H₂SO₄ react in 2:1 molar ratio.

$\frac{0.2275}{0.08972}>2$, hence H₂SO₄ is the limiting reactant.

Now the theoretical yield of Na₂SO₄ can be calculated:

$Theoretical\ yield\ (Na_2SO_4)=0.08972\ mol(H_2SO_4)\times\frac{1\ mol(Na_2SO_4)}{1\ mol(H_2SO_4)}\times\frac{142.04\ g(Na_2SO_4)}{1\ mol(Na_2SO_4)}=12.74\ g$

$\%\ yield=\frac{actual\ yield}{theoretical\ yield}\times100\%=\frac{10.6\ g}{12.74\ g}\times100\%=83\%$

Answer: 83%