Answer :
Answer:
Theoretical yield = 8.55 g
Explanation:
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
For [tex]H_2SO_4[/tex]
Mass of water = 5.9 g
Molar mass of water = 98.079 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{5.9\ g}{98.079\ g/mol}[/tex]
[tex]Moles\ of\ Sulfuric\ acid= 0.0602\ mol[/tex]
Given: For [tex]NaOH[/tex]
Given mass = 6.6 g
Molar mass of NaOH = 39.997 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{6.6\ g}{39.997\ g/mol}[/tex]
[tex]Moles\ of\ NaOH= 0.1650\ mol[/tex]
According to the given reaction:
[tex]2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O[/tex]
1 mole of sulfuric acid reacts with 2 moles of NaOH
So,
0.0602 mole of sulfuric acid reacts with 2*0.0602 moles of NaOH
Moles of NaOH = 0.1204 moles
Available moles of [tex]NaOH[/tex] = 0.1650 moles
Limiting reagent is the one which is present in small amount. Thus, [tex]H_2SO_4[/tex] is limiting reagent.
The formation of the product is governed by the limiting reagent. So,
1 mole of sulfuric acid produces 1 mole of sodium sulfate
So,
0.0602 mole of sulfuric acid produces 0.0602 mole of sodium sulfate
Moles of sodium sulfate = 0.0602 mole
Molar mass of sodium sulfate = 142.04 g/mol
Mass of sodium sulfate = Moles × Molar mass = 0.0602 × 142.04 g = 8.55 g
Theoretical yield = 8.55 g