Asked by auro
determine the quantity of NaOh whose percentage is 15 % to balance 176 g of H2SO4
Answers
Answered by
DrBob222
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.
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.
Answered by
auro
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 ?
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 ?
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