Answer:
The amount of sodium chloride formed is 0.03 moles which are contained in 1.75 g
Explanation:
The reaction is this:
BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl
5g 3g 3.5g x
First of all, let's calculate the moles of everything
Mol = Mass / Molar mass
5g / 208.23 g/m = 0.024 mol BaCl₂
3g / 142.06 g/m = 0.0211 mol Na₂SO₄
3.5g / 233.39 g/m = 0.0150 mol BaSO₄
We can now determinate, the limiting reactant to work
Ratio between reactants is 1:1, so 1 mol of chloride reacts with 1 mol of sulfate. Limiting agent is the Na₂SO₄ ( I need 0.024 mol to react the BaCl₂, and I only have 0.0211 mol, BaCl₂ will remain without reaction)
Ratio is 1:1 between the sulfates so, 0.0211 mol of sodium sulfate produce 0.0211 mol of barium sulfate but I only produced 0.0150 mole of it.
This reaction has a % of yield.
0.0211 moles ___ 100 %
0.0150 moles ___ (0.0150 . 100)/0.0211 = 71.09 %
Now we can know the production of NaCl
Ratio is 1:2, so If i have 0.0211 mol of sodium sulfate I'll produce the double of NaCl (0.0211 .2) = 0.0422 mol
As the reaction has a 71.09% yield, I'll produce
0.0422 mol . 0.7109 = 0.03mol
Molar mass NaCl = 58.45 g/m
0.03 m . 58.45 g/m = 1.75 g
