In order to find the mass of the precipitate formed by the reaction, use the fact that you have a #color(purple)(2):1# mole ratio between sodium phosphate and copper(II) phosphate. More specifically, the reaction will consume This means that all the moles of sodium phosphate present in your sample will take part in the reaction. Now, you know that #'0.659 g'# of sodium phosphate reacts with excess copper(II) sulfate. In order to balance this chemical equation, you must add a coefficient of #color(blue)(3)# to copper(II) sulfate and sodium sulfate a coefficient of #color(purple)(2)# to sodium phosphate. The down arrow added next to copper(II) phosphate indicates that this compound precipitates out of solution. The unbalanced chemical equation that describes this reaction looks like this The second product is aqueous sodium sulfate, #'Na'_ 2'SO'_4#.

This is what a precipitate actually is - an insoluble ionic compound that 'falls' out of solution, i.e.

The copper(II) cations will combine with the phosphate anions to form copper(II) phosphate, #'Cu'_ 3('PO'_4)_2#, an insoluble ionic compound that precipitates out of solution. #'Na'_ 3'PO'_ (4(aq)) -> 3'Na'_ ((aq))^(+) + 'PO'_ (4(aq))^(3-)#Īt this point, you should be familiar with the solubility rules for ionic compounds. Now, both reactants are soluble ionic compounds, which means that they dissociate completely in aqueous solution to form cations and anions. You know that copper(II) sulfate reacts with sodium phosphate, so you can say that

When a monograph specifies that an article responds to the test for dry chlorides, mix the solid to be tested with an equal weight of manganese dioxide, moisten with sulfuric acid, and gently heat the mixture: chlorine, which is recognizable by the production of a blue color with moistened starch iodide paper, is evolved.The first thing to do here is to predict the products and write an unbalanced chemical equation that describes this double replacement reaction. To convert this to grams, use the molar mass of copper (II) phosphate. Add ammonia TS dropwise to this precipitate. In order to find the mass of the precipitate formed by the reaction, use the fact that you have a 2:1 mole ratio between sodium phosphate and copper (II) phosphate. Wash the precipitate with three 1-mL portions of nitric acid solution (1 in 100), and discard the washings. Centrifuge the mixture without delay, and decant the supernatant layer. When testing amine (including alkaloidal) hydrochlorides that do not respond to the above test, add one drop of diluted nitric acid and 0.5 mL of silver nitrate TS to a solution of the substance being examined containing, unless otherwise directed in the monograph, about 2 mg of chloride ion in 2 mL: a white, curdy precipitate is formed. With silver nitrate TS, solutions of chlorides yield a white, curdy precipitate that is insoluble in nitric acid but is soluble in a slight excess of 6 N ammonium hydroxide.

Na3po4 and nacl precipitate color