M(BaCl2)=100 г
+H2SO4
m(осаду)-?
BaCl2+H2SO4->BaSO4+2HCL

Предыдущий
вопрос Следующий
вопрос

1 моль BaCl2 reacts with 1 mole H2SO4 to produce 1 mole BaSO4.

The molar mass of BaCl2 is 208.23 g/mol.
From the given mass of BaCl2 (100 g), we can calculate the number of moles:
100 g / 208.23 g/mol = 0.48 mol

Since the reaction is 1:1 between BaCl2 and H2SO4, the number of moles of H2SO4 required would also be 0.48 mol.

Now, we need to consider the stoichiometry of the reaction:
BaCl2 + H2SO4 -> BaSO4 + 2HCl

There are 2 moles of HCl produced for every 1 mole of BaSO4. Therefore, in the reaction, 0.48 mol of BaSO4 will produce 2 x 0.48 = 0.96 mol of HCl.

Now, we need to calculate the mass of the precipitate (BaSO4):
The molar mass of BaSO4 is 233.39 g/mol.
0.48 mol x 233.39 g/mol = 112.15 g

Therefore, the mass of the precipitate formed (BaSO4) in the reaction is 112.15 g.