Respuesta :
Answer:
514.53 g.
Explanation:
- The balanced equation of the reaction is:
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O.
- It is clear that 2.0 moles of NaOH reacts with 1.0 mole of H₂SO₄ to produce 1.0 mole of Na₂SO₄ and 2.0 moles of H₂O.
- The limiting reactant is NaOH and H₂SO₄ is an unlimited reactant.
- We can calculate the no. of moles of H₂SO₄ using the data available:
n = mass / molar mass = (355.0 g) / (98.0 g/mol) = 3.62 mol.
Using cross multiplication; we can calculate the no. of moles of sodium sulfate Na₂SO₄ can the reaction produce:
1.0 mole of H₂SO₄ produces → 1.0 mole of Na₂SO₄, from the stichiometry.
3.62 moles of H₂SO₄ produces → ??? mole of Na₂SO₄.
- The no. of moles of sodium sulfate Na₂SO₄ can the reaction produce = (1.0 mol)(3.62 mol) / (1.0 mol) = 3.62 mol.
- Now, we can calculate the grams (mass) of sodium sulfate Na₂SO₄ can the reaction produce = n x molar mass = (3.62 mol)(142.04 g/mol) = 514.53 g.
Answer : The theoretical mass of sodium sulfate produced is, 514.04 grams
Solution : Given,
Mass of sulfuric acid = 355 g
Molar mass sulfuric acid = 98 g/mole
Molar mass of sodium sulfate = 142 g/mole
First we have to calculate the moles of sulfuric acid.
[tex]\text{Moles of }H_2SO_4=\frac{\text{Mass of }H_2SO_4}{\text{Molar mass of }H_2SO_4}=\frac{355g}{98/mole}=3.62moles[/tex]
Now we have to calculate the moles of sodium sulfate.
The given balanced reaction is,
[tex]2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O[/tex]
From the balanced reaction, we conclude that
As, 1 mole of sulfuric acid react to give 1 mole of sodium sulfate
So, 3.62 mole of sulfuric acid react to give 3.62 mole of sodium sulfate
Now we have to calculate the theoretical mass of sodium sulfate.
[tex]\text{Mass of }Na_2SO_4=\text{Moles of }Na_2SO_4\times \text{Molar mass of }Na_2SO_4}[/tex]
[tex]\text{Mass of }Na_2SO_4=3.62mole\times 142g/mole=514.04g[/tex]
Therefore, the theoretical mass of sodium sulfate produced is, 514.04 grams
