How many grams of nickel(II)phosphate (366 g/mol)will precipitate if excess sodium phosphate is added to 1.20 L of 0.684 M nickel acetate?

Question

Quizplus · Accepted Answer

The balanced chemical equation is:Ni(CH3COO)2(aq) + Na3PO4(aq) → Ni3(PO4)2(s) + 2CH3COONa(aq)From the stoichiometry of the balanced chemical equation, we can see that 1 mole of Ni(CH3COO)2 reacts with 1 mole of Na3PO4 to produce 1 mole of Ni3(PO4)2.Therefore, the number of moles of Ni3(PO4)2 that will precipitate is equal to the number of moles of Ni(CH3COO)2 in the solution.The number of moles of Ni(CH3COO)2 in the solution is:moles of Ni(CH3COO)2 = concentration of Ni(CH3COO)2 × volume of solution= 0.684 M × 1.20 L= 0.8208 molTherefore, the number of moles of Ni3(PO4)2 that will precipitate is also 0.8208 mol.The mass of Ni3(PO4)2 that will precipitate is:mass of Ni3(PO4)2 = number of moles of Ni3(PO4)2 × molar mass of Ni3(PO4)2= 0.8208 mol × 366 g/mol= 300.0 gTherefore, the correct answer is A) 1.00 × 10^2 g.

How Many Grams of Nickel(II)phosphate (366 G/mol)will Precipitate If Excess ×\times×

How Many Grams of Nickel(II)phosphate (366 G/mol)will Precipitate If Excess $\times$