Four-part question set: A chemist dissolves 18.0 g of cobalt(ll) nitrate in one

Question

Four-part question set: A chemist dissolves 18.0 g of cobalt(ll) nitrate in one beaker of water and 12.0 g of sodium phosphate in a second beaker of water. When the two solutions are poured together, a solid and predict the expected mass of solid product. (Ignore equilibrium.) 28. What is the identity of the solid product? 29. Add together all the coefficients in the balanced equation. What is the total? 30. What is the identity of the limiting reactant? 31. What is the predicted mass of the solid product? 28 29 30. 31.

Explanation / Answer

28. Balanced chemical reaction is

3 Co(NO3)2(aq) + 2 Na3PO4(aq) gives Co3(PO4)2(s) + 6 NaNO3(aq)

solid product is Co3(PO4)2

29. 3 Co(NO3)2(aq) + 2 Na3PO4(aq) gives Co3(PO4)2(s) + 6 NaNO3(aq)

30. As per balnced chemical equation one mole of Na3PO4 reacts with 1.5 mole Co(NO3)2

Number of moles of Co(NO3)2 = mass/molar mass = 18/182.943 = 0.0983

Number of moles of Na3PO4= mass/molar mass = 12/163.94 = 0.0731

1.5 equivalents of Na3PO4 is 0.1097 moles Co(NO3)2. Here we have only 0.0983 moles Co(NO3)2. Thus Co(NO3)2 is the limiting reagent.

31. As per the balnced chemical equation one mole Co(NO3)2 produces 0.333 moles product.

thus maximum theoretical yield is = 0.333 x 0.0983 = 0.0327 moles

molar mass of cobalt phosphate = 366.7423 g/mole

theoretical mass of cobalt phosphate = number of moles x molar mass = 366.7423 x 0.0327 = 12.004 g

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