if a solution containing 50.186G of mercury(ll) nitrate is allowed to react comp
ID: 481878 • Letter: I
Question
if a solution containing 50.186G of mercury(ll) nitrate is allowed to react completely with a solution containing 15.488G of sodium dichromate how many grams of solid precipitate will be formed Question 4 of 32 noottect General Chemistry 4th Edition University Science Books prasented by Sapling Leaming If a solution 50, 186 g of m with a solution II) nitrate is allowed to react containing 15.488 g of sodium dichromate how many grams of solid precipitate wil be formed? Number How many grams of the reactant in excess will remain after the reaction? Number 31.16 is hint available Viuw the hint by ducking on the Click on the bottom divioer divider Bar again Hide hinlExplanation / Answer
2 Hg(NO3)2 + 2 Na2Cr2O7 -----------> Hg2(Cr2O7)2 (ppt) + 4 NaNO3
Find the limiting agent ,
50.186 gm * (2mol /324.7 g Hg(NO3)2) * (1 mol of ppt /2mol Hg(NO3)2) * (833.156 g ppt / 1mol ppt)
= 128.7735 gm of Hg2(Cr2O7)2
15.488 gm * (2mol /261.97 g Na2Cr2O7) * (1molppt/2mol Na2Cr2O7) * (833.156 g ppt / 1mol ppt)
= 49.2572 gm of Hg2(Cr2O7)2
Na2Cr2O7 is the limiting agent, 49.2572 gms of solid precipitate is formed
mol of Na2Cr2O7 = 15.488 gm / 261.97 g/mol = 0.059121 mol Na2Cr2O7
mol of excess reactant used up = (0.059121 mol Na2Cr2O7) * (2mol Hg(NO3)2 / 2mol Na2Cr2O7)
mol of excess reactant used up = 0.059121 mol Hg(NO3)2
mass of excess reactant used up = 0.059121mol Hg(NO3)2* 324.7gm/mol = 19.19658 gm Hg(NO3)2
Therefore grams of the reactant in excess that will remain after the reaction,
50.186 gm - 19.19658 gm = 30.98941 gms
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