determine the quantity of NaOh whose percentage is 15 % to balance 176 g of H2SO4

Do mean to calculate the volume of NaOH required for the neutralization?

2NaOH + H2SO4 ==> Na2SO4 + 2H2O

moles H2SO4 = grams/molar mass
Use the coefficients in the balanced equation to convert moles H2SO4 to moles NaOH.
15% NaOH means 15 g/100 mL (is that 15% w/v?)
So 15 g NaOH = ?? moles.
moles/0.1 L = M
Then M NaOH = moles NaOH/L NaOH
Substitute MNaOH, moles from above of NaOH needed, solve for L NaOH needed.
Post your work if you get stuck.

i need to calculate the mass (m ) of NaoH with C% = 15 % , to balance 176 g of H2SO4

so i though to use this method
C% = mass of h2s04 / mass of h2so4 + mass of Naoh which is unknown

mass of Naoh = c% x mass of h2so4 / 100

is this the right method or not ?

To determine the quantity of NaOH needed to balance 176 g of H2SO4, we first need to write a balanced chemical equation for the reaction between NaOH and H2SO4.

The balanced chemical equation for this reaction is:

2NaOH + H2SO4 -> Na2SO4 + 2H2O

From the balanced equation, we can see that 2 moles of NaOH react with 1 mole of H2SO4.

To calculate the moles of H2SO4, we need to divide the given mass (176 g) by its molar mass. The molar mass of H2SO4 is calculated as follows:

H = 1.01 g/mol (2 atoms)
S = 32.07 g/mol (1 atom)
O = 16.00 g/mol (4 atoms)

Molar mass of H2SO4 = (2 * 1.01) + 32.07 + (4 * 16.00) = 98.09 g/mol

Now, let's calculate the moles of H2SO4:

Moles = Mass / Molar mass
Moles = 176 g / 98.09 g/mol ≈ 1.79 mol

Since the mole ratio between NaOH and H2SO4 is 2:1, we can determine the moles of NaOH required:

Moles of NaOH = 2 * Moles of H2SO4
Moles of NaOH = 2 * 1.79 mol = 3.58 mol

Now, to determine the quantity of NaOH in grams, we need to use the molar mass of NaOH, which is:

Na = 22.99 g/mol (1 atom)
O = 16.00 g/mol (1 atom)
H = 1.01 g/mol (1 atom)

Molar mass of NaOH = 22.99 + 16.00 + 1.01 = 40.00 g/mol

Finally, let's calculate the quantity of NaOH:

Mass = Moles * Molar mass
Mass = 3.58 mol * 40.00 g/mol ≈ 143.20 g

Therefore, approximately 143.20 grams of NaOH with a percentage of 15% is needed to balance 176 grams of H2SO4.