Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of sodium sulfate is produced from the reaction of of sulfuric acid and of sodium hydroxide, calculate the percent yield of sodium sulfate. Be sure your answer has the correct number of significant digits in it.

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

[tex]2NaOH+H_2SO_4-->Na_2SO_4+2H_2O[/tex]

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

[tex]n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\[/tex]

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

[tex]0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4[/tex]

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

[tex]0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4[/tex]

Finally, the percent yield turns out into:

[tex]Y=\frac{1.92g}{7.1g} *100[/tex]

[tex]Y=27.0[/tex]%

Best regards.