Answer:
160 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 [tex]H_2SO_4[/tex] = 196 g
Molar mass of [tex]H_2SO_4[/tex] = 98 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{196\ g}{98\ g/mol}[/tex]
[tex]Moles\ of\ Sulfuric\ acid= 2\ 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,
2 moles of sulfuric acid reacts with 2*2 moles of NaOH
Moles of NaOH must react = 4 moles
Molar mass of NaOH = 40 g/mol
Mass = Moles*molar mass = [tex]4\times 40\ g[/tex] = 160 g
Final answer:
To react completely with 196 grams of H₂SO₄, you would need 80 grams of NaOH. This calculation involves balancing the chemical equation and using stoichiometry.
Explanation:
Step 1: Write and balance the chemical equation:
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
Step 2: Calculate the molar mass of NaOH (40 g/mol) and H₂SO₄ (98 g/mol).
Step 3: Use stoichiometry to find the moles of NaOH needed to react completely with 196 grams of H₂SO₄. From the balanced equation, 1 mole of H₂SO₄ reacts with 2 moles of NaOH.
Step 4: Calculate the total number of grams of NaOH needed:
(196 g H₂SO₄) x (1 mol NaOH / 98 g H₂SO₄) x (40 g NaOH / 1 mol NaOH) = 80 grams of NaOH
A student preparing for the experiments inadvertently adds an additional 400 mL of the same acid solution to the dissolution vessel. What will be the new pOH of this solution?
Answer:
POH= 13
Explanation:
A student preparing for the experiments inadvertently adds an additional 400 mL of the same acid solution to the dissolution vessel. What will be the new pOH of this solution?
PH is the measure of the degree of acidity of a solution.
POH is the measure of the degree of alkalinity of a solution
Note that pH + pOH = 14
if concentration remains the same, then volume changes will not affect pH.
The pH of the solution is given as
PH= -log[H+].
For this experiment, the dissolution vessel contains 0.1 M HCl, no matter the initial volume of the acid solution
For the molar concentration of the cation, we can propose that a strong acid will dissociate completely,
[H+] = 0.1 = 1 x 10-1 M
. substituting the concentration of the cation
PH=-log[1 x 10-1] = 1.
Note that pH + pOH = 14 for any aqueous solution.
we say that
the pOH = 14 - pH.
pOH = 14-1 = 13.