Ask Question
17 January, 14:17

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

+2
Answers (1)
  1. 17 January, 15:18
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “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 ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers