A 190.0 mL solution of 2.613 M strontium nitrate is mixed with 220.0 ml of a 3.3

Question

A 190.0 mL solution of 2.613 M strontium nitrate is mixed with 220.0 ml of a 3.301 M sodium fluoride solution. Calculate the mass of the resulting strontium fluoride precipitate. Assuming complete precipitation, calculate the final concentration of each ion. If the ion is no longer in solution, enter a zero for the concentration. How many grams of a 11.9% sugar solution contain 21.0 g of sugar? If a solution containing 98.867 g of mercury(ll) perchlorate is allowed to react completely with a solution containing 17.796 g ot sodium sulfide, how many grams of solid precipitate will be formed? How many grams of the reactant in excess will remain after the reaction?

Explanation / Answer

only 1 st should be answered.As it was marked 1 st iam doing fom this question

(1)

100 grams of sugar solution contains 11.9 grams of sugar

but here we required 21 grams of sugar.

So, weight of sugar solution == (21 grams of sugar / 11.9 grams of sugar) x 100 grams of sugar solution

=176.47 grams of sugar solution

(2)

The reaction is as follows:

Hg(O2CCH3)2 + Na2S -------->>> HgS(s) + 2 NaO2CCH3(aq)

From the reaction it is clear that 1 mole Hg(C2O2H3)2 reacts with 1mole Na2S to produce 1 mole HgS which is the solid precipitate

98.867 g= 98.867/318.68=0.310 moles Hg(C2O2H3)2

17.796 g Na2S=7.41/78=0.228 moles

So Na2S is the limiting reagent

So 0.228 moles Na2S react with 0.228 moles Hg(C2O2H3)2 to produce 0.228 moles HgS

Mass of HgS produced=0.228x232.66=53.08 g

Excess reactant=(0.310-0.228) moles Hg(C2O2H3)2

=0.082x318.68 g

=26.13 g

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