A mixture, weighing 6.000 grams, contains 41.50 % Na₃PO₄ and 58.50 % BaCl₂ of the mixture, by weight. When dissolved in water, a precipitate forms via the double displacement reaction:BaCl₂(aq)+ Na₃PO₄(aq) NaCl(aq)+ Ba₃(PO₄)₂(s)If the reaction is quantitative, the yield of the solid barium phosphate should be ,Hint: You must balance the equation first then determine the limiting reagent.

Question

Quizplus · Accepted Answer

The balanced equation is:3BaCl2(aq) + 2Na3PO4(aq) → 6NaCl(aq) + Ba3(PO4)2(s)The molar mass of Na3PO4 is 163.94 g/mol, and the molar mass of BaCl2 is 208.23 g/mol.The number of moles of Na3PO4 in the mixture is:(41.50 g Na3PO4) / (163.94 g/mol Na3PO4) = 0.253 mol Na3PO4The number of moles of BaCl2 in the mixture is:(58.50 g BaCl2) / (208.23 g/mol BaCl2) = 0.281 mol BaCl2The mole ratio of BaCl2 to Na3PO4 is 3:2, so BaCl2 is the limiting reagent.The theoretical yield of Ba3(PO4)2 is:(0.281 mol BaCl2) × (1 mol Ba3(PO4)2 / 3 mol BaCl2) × (601.98 g/mol Ba3(PO4)2) = 3.382 g Ba3(PO4)2Therefore, the correct answer is A: 3.382 grams.

A Mixture, Weighing 6