A 0.500 g sample of a BaCl2•2H2O/Na2SO4 salt mixture, when mixed with water,
filtered, and dried, produced 0.225 g of BaSO4. The Na2SO4 was determined to be
the limiting reactant.
calculate the % of Na2So4 and BaCl2.2hro in the salt solution
Convert grams BaSO4 ppt to g Na2SO4.
moles BaSO4 = grams/molar mass
moles Na2SO4 = moles BaSO4.
grams Na2SO4 = moles Na2SO4 x molar mass Na2SO4.
%Na2SO4 = (mass Na2SO4/mass sample)*100=?
To determine the mass of Na2SO4 in the salt mixture, we need to calculate the amount of BaSO4 that can be formed from the reaction. Since Na2SO4 is the limiting reactant, it determines the maximum amount of BaSO4 that can be obtained.
The molar mass of BaSO4 is:
Ba = 137.33 g/mol
S = 32.07 g/mol
O = 16.00 g/mol
Total = 137.33 + 32.07 + (16.00 * 4) = 233.38 g/mol
Now, let's calculate the number of moles of BaSO4 that can be formed:
moles of BaSO4 = mass of BaSO4 / molar mass of BaSO4
moles of BaSO4 = 0.225 g / 233.38 g/mol
moles of BaSO4 = 0.000963 mol
From the balanced equation, we can see that one mole of BaSO4 is formed from one mole of Na2SO4. Therefore, the number of moles of Na2SO4 in the mixture is also 0.000963 mol.
To convert this to grams, we can use the molar mass of Na2SO4 (142.04 g/mol):
mass of Na2SO4 = moles of Na2SO4 * molar mass of Na2SO4
mass of Na2SO4 = 0.000963 mol * 142.04 g/mol
mass of Na2SO4 = 0.137 g
So, the mass of Na2SO4 in the BaCl2•2H2O/Na2SO4 salt mixture is 0.137 g.