25cm3 of 0.1 mole H2SO4 neutralizes 20.0cm3 of impure NaOH containing 3.0g of impure NaOH in 250cm3 Of a solution . calculate the percentage purity of the NaOH (Na=23,O=16,H=1)
2NaOH + H2SO4 ==> Na2SO4 + 2H2O
mols H2SO4 = M x L 0.1M x 0.025 L = 0.0025 mols.
mols NaOH neutralized is 2 x that = 0.0050
That's the amount of pure NaOH in the 20 cc. So how much is in the 250 cc container? That will be 0.005 x (250/20) = ? mols NaOH
grams NaOH in the 3.0 g sample is ?mols NaOH from above x 40 g/mol = ?
The %purity = (grams pure NaOH/3.0)*100 = ?
I get something close to 80% purity.
To calculate the percentage purity of NaOH, we need to determine the number of moles of NaOH and the theoretical number of moles that should have been present if the NaOH was pure.
Let's begin by finding the number of moles of H2SO4:
Moles of H2SO4 = volume (in dm3) × concentration (in mol/dm3)
Moles of H2SO4 = 25 cm3 × (1 dm3/1000 cm3) × 0.1 mol/dm3
Moles of H2SO4 = 0.0025 mol
Since H2SO4 and NaOH react in a 1:2 ratio, we can determine the number of moles of NaOH based on the moles of H2SO4:
Moles of NaOH = 2 × Moles of H2SO4
Moles of NaOH = 2 × 0.0025 mol
Moles of NaOH = 0.005 mol
Now, let's calculate the number of moles of NaOH present in the impure sample:
Moles of NaOH in impure sample = mass (in g) / molar mass (in g/mol)
Moles of NaOH in impure sample = 3.0 g / (23 g/mol + 16 g/mol + 1 g/mol)
Moles of NaOH in impure sample = 3.0 g / 40 g/mol
Moles of NaOH in impure sample = 0.075 mol
Now, we can calculate the percentage purity of NaOH:
Percentage purity = (Moles of pure NaOH / Moles of impure NaOH) × 100%
Percentage purity = (0.005 mol / 0.075 mol) × 100%
Percentage purity = 6.67%
Therefore, the percentage purity of the NaOH is approximately 6.67%.
To calculate the percentage purity of NaOH in the solution, we need to determine the number of moles of NaOH using the given information.
First, let's find the number of moles of H2SO4 using the given volume and concentration:
Volume of H2SO4 = 25 cm^3
Concentration of H2SO4 = 0.1 moles/cm^3
Number of moles of H2SO4 = Volume x Concentration
= 25 cm^3 x 0.1 moles/cm^3
= 2.5 moles
According to the balanced chemical equation for the neutralization reaction between H2SO4 and NaOH, the mole ratio is 1:2. This means that 1 mole of H2SO4 reacts with 2 moles of NaOH.
Since the number of moles of H2SO4 is 2.5, we can calculate the number of moles of NaOH:
Number of moles of NaOH = (2.5 moles H2SO4) * (2 moles NaOH / 1 mole H2SO4)
= 5 moles
Now, let's calculate the number of moles of NaOH in the impure sample using the given mass:
Mass of impure NaOH = 3.0 g
To convert grams to moles, we need to know the molar mass of NaOH:
Molar mass of NaOH = (23 g/mol) + (16 g/mol) + (1 g/mol)
= 40 g/mol
Number of moles of NaOH = Mass / Molar mass
= 3.0 g / 40 g/mol
= 0.075 moles
Finally, we can calculate the percentage purity of NaOH in the solution:
Percentage purity = (Number of moles of pure NaOH / Number of moles of impure NaOH) x 100
Number of moles of pure NaOH = 5 moles
Number of moles of impure NaOH = 0.075 moles
Percentage purity = (5 moles / 0.075 moles) x 100
= 6666.67%
Therefore, the percentage purity of NaOH in the impure sample is approximately 6666.67%.