A 0.5g of fuming h2so4 oleum is diluted with water. This solution is completedly neutralized by 26.7ml of 0.4N naoh. Find the percentage of free so3 in the sample
The asnwer 20.6%
My calculation
h2so4= 98/2 = 49
so3= 80/2 = 40
26.7ml x 0.4N = 10.68ml - 0.01068L
x/49 + x/40 (0.5-X) Bob helps me
Eq. H2SO4 = Eq of NaOH
= (26.7 x 0.4)/1000
= 10.68 mEq.
H2SO4 + SO3 + H2O -----> 2(H2SO4)
>---------< is oleum.
==> SO3 + H2SO4 -----> H2SO4
∴Eq of SO3 = 1/2 of mEq. total H2SO4
==>mEq. of Oleum = Eq of H2SO4 + Eq of SO3
= Eq of total H2SO4
= 10.68/1000 (bec. 1 mEq. = 1/1000 Eq.)
Let x be mass of SO2 in Oleum. ==> mass of H2SO4 = (0.5-x)
Eq. of Oleum = (0.5 - x)/49 + x/40 = 10.68/1000
Solving.... x = 0.1836g
Percentage = 0.1836/0.5 x 100 = 20.73%
Ur Tu
To find the percentage of free SO3 in the sample, you can start by calculating the number of moles of H2SO4 (oleum) and NaOH used in the neutralization reaction.
Given that the weight of the H2SO4 (oleum) is 0.5g and its molar mass is 98 g/mol, you can calculate the number of moles of H2SO4:
moles H2SO4 = mass / molar mass
moles H2SO4 = 0.5g / 98 g/mol
moles H2SO4 = 0.0051 mol
Next, you have the volume and the concentration of NaOH used in the neutralization reaction. The volume is given as 26.7 mL, which can be converted to liters:
volume NaOH = 26.7 mL = 26.7 / 1000 L
volume NaOH = 0.0267 L
The concentration of NaOH is given as 0.4 N, which means it is 0.4 moles per liter. Multiply the concentration by the volume to get moles of NaOH used:
moles NaOH = concentration NaOH x volume NaOH
moles NaOH = 0.4 mol/L x 0.0267 L
moles NaOH = 0.0107 mol
Since the reaction is a 1:1 stoichiometric ratio between H2SO4 and NaOH, the moles of NaOH used represent the moles of H2SO4 neutralized.
Now, we need to find the moles of SO3 present in the sample. The molar mass of SO3 is 80 g/mol. Let's assume that the mass of SO3 is x grams:
moles SO3 = mass SO3 / molar mass SO3
moles SO3 = x g / 80 g/mol
From the given information, we know that the amount of H2SO4 remaining is 0.0051 mol - 0.0107 mol:
moles H2SO4 remaining = moles H2SO4 initial - moles NaOH
moles H2SO4 remaining = 0.0051 mol - 0.0107 mol
moles H2SO4 remaining = -0.0056 mol
Since the reaction goes to completion, all the H2SO4 will be neutralized, and there is no H2SO4 remaining. Therefore, the moles of H2SO4 remaining should be zero. We can create an equation with the moles of SO3 and the moles of H2SO4 remaining:
0 = x g / 80 g/mol - 0.0056 mol
Solve for x:
x g / 80 g/mol = 0.0056 mol
x g = 0.0056 mol x 80 g/mol
x g = 0.448 g
Now, to calculate the percentage of free SO3 in the sample, divide the mass of SO3 by the mass of the original sample (0.5g) and multiply by 100:
percentage of free SO3 = (mass SO3 / mass sample) x 100
percentage of free SO3 = (0.448 g / 0.5 g) x 100
percentage of free SO3 = 0.896 x 100
percentage of free SO3 = 89.6%
Therefore, the percentage of free SO3 in the sample is 89.6%.
To calculate the percentage of free SO3 in the sample, we need to use the information given about the neutralization reaction.
Let's break down the steps to find the solution:
1. Calculate the number of moles of NaOH used:
We are given that 26.7 mL of 0.4 N NaOH was used. N stands for the normality of the solution, which is the number of equivalents per liter. So, 0.4 N means 0.4 equivalents per liter.
To convert the volume to liters, divide by 1000: 26.7 mL / 1000 = 0.0267 L.
To calculate the number of moles, multiply the volume by the normality: 0.0267 L * 0.4 N = 0.01068 moles.
2. Calculate the number of moles of H2SO4:
Since we are given the mass of H2SO4 (0.5 g), we can convert it to moles using the molar mass of H2SO4: 98 g/mol.
To calculate the number of moles, divide the mass by the molar mass: 0.5 g / 98 g/mol = 0.0051 moles.
3. Calculate the number of moles of SO3:
For every 2 moles of H2SO4, there is 1 mole of SO3.
So, the number of moles of SO3 will be half the number of moles of H2SO4: 0.0051 moles / 2 = 0.00255 moles.
4. Calculate the percentage of free SO3:
To find the percentage of free SO3, we need to compare the moles of free SO3 (0.00255 moles) to the moles of H2SO4 used (0.0051 moles).
Divide the moles of free SO3 by the moles of H2SO4 and multiply by 100 to get the percentage: (0.00255 moles / 0.0051 moles) * 100 = 50%.
Therefore, the percentage of free SO3 in the sample is 50%.
It seems there might be a discrepancy between the provided answer (20.6%) and the calculations. Could you please provide more information or clarify any possible errors?