Question

13.33 cg of barium chloride is reacted with 135.2 mL of a 0.250 M sulfuric acid.

Write the Balanced Equation.

What kind of reaction is this?

How many grams of product? (List Product and Amount)

Answer 1

the reaction:

BaCl2(aq) + H2SO4(aq) --> BaSO4(s) + HCl(aq)

balance:

BaCl2(aq) + H2SO4(aq) --> BaSO4(s) + 2HCl(aq)

this reaction must be precipitation reaction, since BaSO4 is not soluble and will form a precipitant

c)

find mass of products

MW BaCl2 = 208.23 g/mol

mol of BaCl2 = mass/MW = 13.33/208.23 = 0.06401 mol of BaCl2

for..

mol of acid:

mol = MV = 135.2*0.25 = 33.8 mmol = 33.8*10^-3 mol = 0.0338 mol of H2SO4

so...

mol of Ba+2 ions = 0.06401

mol of SO4-2 ions = 0.0338

so...

limiting reactant is clearly SO4-2, since 0.0338 mol of SO4-2 requires 0.0338 mol of Ba+2

so...

mass of BaSO4 formed --> 0.0338 mol

mass = mol*MW = 0.0338 *233.43 = 7.889934 g of BaSO4