Help me answer this!!

Question

You have reacted 100.00 mL of 0.934M calcium nitrate solution with 100.00 mL of 1.051M sodium phosphate solution. After filtering and drying the precipitate, you find that you have 7.473g of calcium phosphate solid. What is your percent yield? Report your answer to 1 decimal place.

Answer 1

Once again Jena, I answered this question FROM YOU several days ago. The only difference between this question and the original one (also answered by @ John T) are some of the molarities used. The procedure for answering is the same. I strongly suggest you go back and look at our answers and try to do this problem yourself. No one will be around to assist you when you take a test on this material.

Answer 2

Hi Jena A.,

You should have a systematic approach to all such questions. They may require a specific chemical knowledge, but they also usually require writing a balanced chemical equation for a reaction, and transforming amounts of materials back and forth from units of moles to units of mass.

So what do you need here?

First of all, you need to write the reaction of Ca(²⁺) with (PO₄)(^3-). Write that using as sources of the ions the reactants specified, namely Ca(NO₃)₂ and Na₃PO₄ . Balance it (make sure that all numbers of each element of atoms are balanced on each side of the equation). Here is where specific chemical knowledge is required: you should have learned (by now) that phosphate salts of Group 2 elements are insoluble! (by insoluble, I mean they form M_x(PO₄)_y solids, where x and y are integers. You know that by perhaps having done a lab exercise on carbonate salts of Group 2 salts, they are variously insoluble, and the sulfates are more insoluble, so that the phosphates will be very insoluble. Higher charge ions == more insoluble!!) Now, look closely at the volumes and concentrations stated for each of those reactants. The volumes and concentrations given (as M) are about equal; that means that the Na₃(PO₄) is in excess, since it would require 1.5 times as much of the calcium salt to balance what you have. So calculate on the basis of the limiting reactant, namely, the calcium salt, just how much (as #moles) of Ca₃(PO₄)₂ could possibly be made from that amount of calcium salt reactant. (use input moles, ratio through using the coefficients of the balanced equation, to get output moles).

All right, now add up the molar mass for that calcium phosphate product. Convert that #moles above to a mass, and compare to the mass obtained. (divide actual=experimental by theoretical=what you just calculated above). That is a decimal; convert to percent, rounded properly as requested.

Note: carry ALL the precision of the input data through your calculations; round as requested as the very LAST thing you do, to the result.

-- Cheers, -- Mr. d.