8 g of sulphur is burnt to form SO2 which is oxidized by Cl2 water The solution is treated with BaCl2 solution The amount of BaSO4 precipitated is
A/ I mole
B/ 0.5 mole
C/ 0.24 mole
D/. 0.25 mole
Which option is correct??
To find the amount of BaSO4 precipitated, we need to determine the chemical equation for the reaction and calculate the stoichiometry.
The balanced equation for the reaction can be written as follows:
SO2 + Cl2 + H2O -> H2SO4 + HCl
From the equation, we can see that:
1 mole of SO2 reacts with 1 mole of Cl2 and 1 mole of H2O to produce 1 mole of H2SO4 and 1 mole of HCl.
Given that 8 g of sulfur is burnt, we need to calculate the number of moles of SO2 produced.
The molar mass of sulfur (S) is approximately 32 g/mol. Therefore, the number of moles of sulfur is calculated as follows:
Number of moles = mass / molar mass
Number of moles = 8 g / 32 g/mol = 0.25 mol
Since the number of moles of SO2 is the same as the number of moles of BaSO4 precipitated, the correct option is D: 0.25 mole
To determine the amount of BaSO4 precipitated, we need to calculate the moles of BaSO4 formed from the given information.
The balanced equation between SO2 and BaSO4 is:
SO2 + BaCl2 + H2O → BaSO4 + 2HCl
From the equation, we can see that 1 mole of SO2 reacts with 1 mole of BaSO4. Therefore, the moles of BaSO4 formed will be the same as the moles of SO2 used.
Given that 8 g of sulphur is burnt to form SO2, we need to convert the mass of sulphur to moles:
Molar mass of sulfur (S) = 32 g/mol
Moles of sulfur = mass/ molar mass = 8 g/ 32 g/mol = 0.25 mol
Therefore, 0.25 moles of BaSO4 will be formed. So, the correct option is:
D/ 0.25 mole
S + O2 ==> SO2
SO2 + Cl2 ==> SO4^2-
mols S = grams/atomic mass = ?
1 mol S = 1 mol SO4^2-
1 mol SO4^2- = 1 mol BaSO4
I believe that is D