When 5 g of barium chloride reacted with 3 g of sodium sulfate the products formed were 3.5 g of barium sulfate and x gram of sodium chloride what will be the amount of sodium chloride ph5 g of barium chloride reacted with 3 g of sodium sulfate the products formed were 3.5 g of barium sulfate and x gram of sodium chloride what will be the amount of sodium chloride formed

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