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Q4. If 77.00 mL of 1.655 M magnesium sulfate solution is combined with 85.00 mL of...

Q4. If 77.00 mL of 1.655 M magnesium sulfate solution is combined with 85.00 mL of 1.294 M sodium phosphate solution, and 7.647 g of precipitate is recovered, what is the percent yield?

Expert Solution

3MgSO4 + 2Na3PO4 -------> Mg3(PO4)2 + 3Na2SO4

Stoichiometric mass of MgSO4 = 3×120.366g = 361.098g

Stoichiometric mass of Na3PO4 = 2 × 163.94g = 327.88g

Stoichiometric mass of Mg3(PO4)2=1 × 262.857g = 262.857g

given mass of MgSO4 = ((1.655mol/1000ml)×77ml)×120.366g = 15.3388g

given mass of Na3PO4 = (1.294mol/1000ml)×85ml = 18.0317g

stoichiometrically 361.098g of MgSO4 require 327.88g of Na3PO4

Therefore, 15.3388g of MgSO4 require (327.88/361.098)×15.3388g = 13.9277g of Na3PO4

but actual mass of Na3PO4 is 18.0317g

So, MgSO4 is limiting

361.098g of MgSO4 give 262.857g of Mg3(PO4)2

Therefore, 15.3388g of MgSO4 give

(262.857/361.098)×15.3388g = 11.1656g of Mg3(PO4)2

Actual yield = 7.647g

Percent yield =( actual yield / theoretical yield)×100

= (7.647g/11.1657g)×100

= 68.49℅

queen_honey_blossom answered 2 years ago

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