Question

OBSERVATIONS
1. Mass of 50 mL distilled water (assuming density as 1g /cc) = 50.0 g
2. Mass of BaCl, H,0 = 3.6 g
3. Mass of BaCl, solution = 53.6 g
4. Mass of Na2SO4.10H,0 = 8.05 g
5. Mass of Na,so, solution = 58.05 g
6. Total mass of reactants (solutions of BaCl, and Na 80.) = 53.6 + 58.05 = 111.65 g
7. Mass of empty 150 mL beaker, m =........
8. Mass of reaction mixture before precipitation, m2 = m + 111.65 g =
9. Final mass of reaction mixture after precipitation, mg =
вратит
find
in​

Answer 1

Answer:

The molar mass of Na

2

SO

4

.10H

2

O is

2(23)+32+4(16)+10(18)=322 g/mol.

1 mole of Na

2

SO

4

.10H

2

O contains 14 moles of O atoms.

1 mole of Na

2

SO

4

.10H

2

O corresponds to 14×16=224 g of O.

Amount of oxygen in 32.2g of Na

2

SO

4

.10H

2

O is 32.2×224322=22.4 g