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50g of 10% Barium chloride is mixed with 50g of 10% sodium sulphate in a closed vessel after sometime 6.83 gram of Barium sulphate was precipitated point besides that reaction mixture contained 90 gram of water and sodium chloride point what is the amount of sodium chloride formed?
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50g of 10% Barium chloride is mixed with 50g of 10% sodium sulphate in...
119g of NaCl produce= 170 of NaNO3
1g of NaCl produce= 170/119 of NaNO3
so 5g of NaCl produce= 170/119 x 5
=7.143g of NaNO3
This question is part of UPSC exam. View all Class 9 courses
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50g of 10% Barium chloride is mixed with 50g of 10% sodium sulphate in...
Given:
- Mass of Barium chloride (BaCl2) = 50g
- Concentration of Barium chloride = 10%
- Mass of sodium sulphate (Na2SO4) = 50g
- Concentration of sodium sulphate = 10%
- Mass of Barium sulphate (BaSO4) precipitated = 6.83g
- Water content in the reaction mixture = 90g
To Find:
- Amount of sodium chloride (NaCl) formed
Solution:
Step 1: Calculate the moles of Barium chloride (BaCl2)
- The molar mass of BaCl2 = atomic mass of Ba + 2 * atomic mass of Cl = 137.33 + 2 * 35.45 = 208.23 g/mol
- Moles of BaCl2 = mass of BaCl2 / molar mass of BaCl2 = 50g / 208.23 g/mol
- Moles of BaCl2 = 0.24 mol
Step 2: Calculate the moles of sodium sulphate (Na2SO4)
- The molar mass of Na2SO4 = 2 * atomic mass of Na + atomic mass of S + 4 * atomic mass of O = 2 * 22.99 + 32.07 + 4 * 16.00 = 142.04 g/mol
- Moles of Na2SO4 = mass of Na2SO4 / molar mass of Na2SO4 = 50g / 142.04 g/mol
- Moles of Na2SO4 = 0.35 mol
Step 3: Determine the limiting reactant
- The balanced chemical equation for the reaction between BaCl2 and Na2SO4 is:
BaCl2 + Na2SO4 → BaSO4 + 2NaCl
- From the equation, it can be observed that 1 mole of BaCl2 reacts with 1 mole of Na2SO4 to produce 1 mole of BaSO4 and 2 moles of NaCl.
- The molar ratio between BaCl2 and Na2SO4 is 1:1
- Therefore, the limiting reactant is BaCl2 since the moles of BaCl2 (0.24 mol) is less than the moles of Na2SO4 (0.35 mol).
Step 4: Calculate the moles of BaSO4 precipitated
- From the balanced chemical equation, it can be observed that 1 mole of BaCl2 reacts to produce 1 mole of BaSO4.
- Therefore, the moles of BaSO4 = moles of BaCl2 = 0.24 mol
Step 5: Calculate the moles of NaCl formed
- From the balanced chemical equation, it can be observed that 1 mole of BaCl2 reacts to produce 2 moles of NaCl.
- Therefore, the moles of NaCl = 2 * moles of BaCl2 = 2 * 0.24 mol
- Moles of NaCl = 0.48 mol
Step 6: Calculate the mass of NaCl formed
- The m
- Mass of Barium chloride (BaCl2) = 50g
- Concentration of Barium chloride = 10%
- Mass of sodium sulphate (Na2SO4) = 50g
- Concentration of sodium sulphate = 10%
- Mass of Barium sulphate (BaSO4) precipitated = 6.83g
- Water content in the reaction mixture = 90g
To Find:
- Amount of sodium chloride (NaCl) formed
Solution:
Step 1: Calculate the moles of Barium chloride (BaCl2)
- The molar mass of BaCl2 = atomic mass of Ba + 2 * atomic mass of Cl = 137.33 + 2 * 35.45 = 208.23 g/mol
- Moles of BaCl2 = mass of BaCl2 / molar mass of BaCl2 = 50g / 208.23 g/mol
- Moles of BaCl2 = 0.24 mol
Step 2: Calculate the moles of sodium sulphate (Na2SO4)
- The molar mass of Na2SO4 = 2 * atomic mass of Na + atomic mass of S + 4 * atomic mass of O = 2 * 22.99 + 32.07 + 4 * 16.00 = 142.04 g/mol
- Moles of Na2SO4 = mass of Na2SO4 / molar mass of Na2SO4 = 50g / 142.04 g/mol
- Moles of Na2SO4 = 0.35 mol
Step 3: Determine the limiting reactant
- The balanced chemical equation for the reaction between BaCl2 and Na2SO4 is:
BaCl2 + Na2SO4 → BaSO4 + 2NaCl
- From the equation, it can be observed that 1 mole of BaCl2 reacts with 1 mole of Na2SO4 to produce 1 mole of BaSO4 and 2 moles of NaCl.
- The molar ratio between BaCl2 and Na2SO4 is 1:1
- Therefore, the limiting reactant is BaCl2 since the moles of BaCl2 (0.24 mol) is less than the moles of Na2SO4 (0.35 mol).
Step 4: Calculate the moles of BaSO4 precipitated
- From the balanced chemical equation, it can be observed that 1 mole of BaCl2 reacts to produce 1 mole of BaSO4.
- Therefore, the moles of BaSO4 = moles of BaCl2 = 0.24 mol
Step 5: Calculate the moles of NaCl formed
- From the balanced chemical equation, it can be observed that 1 mole of BaCl2 reacts to produce 2 moles of NaCl.
- Therefore, the moles of NaCl = 2 * moles of BaCl2 = 2 * 0.24 mol
- Moles of NaCl = 0.48 mol
Step 6: Calculate the mass of NaCl formed
- The m
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50g of 10% Barium chloride is mixed with 50g of 10% sodium sulphate in a closed vessel after sometime 6.83 gram of Barium sulphate was precipitated point besides that reaction mixture contained 90 gram of water and sodium chloride point what is the amount of sodium chloride formed?
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50g of 10% Barium chloride is mixed with 50g of 10% sodium sulphate in a closed vessel after sometime 6.83 gram of Barium sulphate was precipitated point besides that reaction mixture contained 90 gram of water and sodium chloride point what is the amount of sodium chloride formed? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about 50g of 10% Barium chloride is mixed with 50g of 10% sodium sulphate in a closed vessel after sometime 6.83 gram of Barium sulphate was precipitated point besides that reaction mixture contained 90 gram of water and sodium chloride point what is the amount of sodium chloride formed? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 50g of 10% Barium chloride is mixed with 50g of 10% sodium sulphate in a closed vessel after sometime 6.83 gram of Barium sulphate was precipitated point besides that reaction mixture contained 90 gram of water and sodium chloride point what is the amount of sodium chloride formed?.
50g of 10% Barium chloride is mixed with 50g of 10% sodium sulphate in a closed vessel after sometime 6.83 gram of Barium sulphate was precipitated point besides that reaction mixture contained 90 gram of water and sodium chloride point what is the amount of sodium chloride formed? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about 50g of 10% Barium chloride is mixed with 50g of 10% sodium sulphate in a closed vessel after sometime 6.83 gram of Barium sulphate was precipitated point besides that reaction mixture contained 90 gram of water and sodium chloride point what is the amount of sodium chloride formed? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 50g of 10% Barium chloride is mixed with 50g of 10% sodium sulphate in a closed vessel after sometime 6.83 gram of Barium sulphate was precipitated point besides that reaction mixture contained 90 gram of water and sodium chloride point what is the amount of sodium chloride formed?.
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