0.5g of fuming h2s04 oleum is diluted with water. This solution is completely neutralized by 26.7ml of 0.4N naoh. Find the percentage of free so3 in the sample
My calculation
h2s04 = 98g/mol
s03 = 80g/mol
98/2 = 49
80/2 = 40
26.7ml x 0.4n = 10.68ml - 0.01067L
x/49 + x/40 (0.5-x) = 0.01068L Please help me The answer 20.6%
To find the percentage of free SO3 in the sample, we can start by setting up a balanced equation for the neutralization reaction between H2SO4 and NaOH:
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
From the balanced equation, we can see that 1 mole of H2SO4 reacts with 2 moles of NaOH. Since the concentration of NaOH is given in terms of Normality (N), we need to convert it to moles.
N = mol/L
0.4 N NaOH means that there are 0.4 moles of NaOH in 1 liter of solution. So, in 26.7 mL (0.0267 L) of 0.4N NaOH, the number of moles of NaOH can be calculated as:
0.4 N x 0.0267 L = 0.01068 moles
According to the reaction stoichiometry, this means that 0.01068 moles of NaOH reacted with 0.00534 moles of H2SO4.
We want to find the percentage of free SO3 in the sample. So, let's assume that x moles of SO3 (out of 0.00534 moles) are present in the sample.
Using the molecular weight of H2SO4, we know that 98g of H2SO4 corresponds to 80g of SO3.
98g H2SO4 / 2 moles H2SO4 = 49g H2SO4 / mole H2SO4
80g SO3 / 2 moles SO3 = 40g SO3 / mole SO3
We can write the following equation based on the stoichiometry and moles of reactants:
x moles SO3 * (49g H2SO4 / mole H2SO4) + x moles SO3 * (40g SO3 / mole SO3) + (0.5 - x) moles SO3 * (40g SO3 / mole SO3) = 0.01068 moles
Rearrange the equation and solve for x:
x(49/98 + 40/80 + 40/80) - 40/80 * 0.5 = 0.01068
Simplify the equation:
3x/2 = 0.01068
x = 0.01068 * 2 / 3
x ≈ 0.00712 moles
To find the percentage of free SO3 in the sample, divide the moles of free SO3 (x) by the moles of H2SO4 used in the initial solution (0.00534 moles) and multiply by 100:
(0.00712 moles / 0.00534 moles) * 100 ≈ 133.25%
Therefore, the percentage of free SO3 in the sample is approximately 133.25%.