An excess of sulfuric acid solution was added to a solution weighing 1 kg with a mass fraction of barium chloride of 6.24%

| January 14, 2021 | education

An excess of sulfuric acid solution was added to a solution weighing 1 kg with a mass fraction of barium chloride of 6.24%. Determine the mass of the formed salt

Let us write the equation for the reaction of the interaction of barium chloride with sulfuric acid.
BaCl2 + H2SO4 = BaSO4 + 2HCl.
Determine the mass of barium chloride. We use the following formula.
W = m r.v. / m r-ra * 100%.
Where w is the mass fraction of the compound.
m = 1000 * 0.00624 = 6.24 g.
Consequently, the mass of barium chloride is 6.24 g.
Next, using the reaction equation, we determine the mass of the resulting precipitate. Let’s calculate the molar mass of barium chloride.
M (BaCl2) = 137 + 35.5 * 2 = 208 g / mol.
Determine the molar mass of barium sulfate.
M (BaSO4) = 137 + 32 + 16 * 4 = 233 g / mol.
Let’s determine the mass of the resulting salt.
6.24 g of barium chloride – in g of barium sulfate.
208 g / mol barium chloride – 233 g / mol barium sulfate.
Y = 233 * 6.24 / 208 = 6.99 g.
When solving, it was found that the mass of the barium sulfate salt is 6.99 grams.
Answer: the mass of the obtained salt is 6.99 g.



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