If a solution containing 34.23 g of mercury(ii) acetate is allowed to react completely with a solution containing 5.102 g of sodium sulfide, how many grams of solid precipitate will be formed? mass: 15.21 g how many grams of the reactant in excess will remain after the reaction? mass: 13.39 g assuming complete precipitation, how many moles of each ion remain in solution? if an ion is no longer in solution, enter a zero (0) for the number of moles. hg2+ :

Na2S is the limiting reactant, 0.06537 mole will be fully consumed.  

mercury(II) acetate is in excess, there remains 0.04203 ( 13.39g)

There will be formed 0.06537 mole HgS (15.21g) and 0.13074 mole of N a C 2 H 3 O 2 (10.725g)

Explanation:

Step 1 : Balance the reaction

C 4 H 6 O 4 H g  +  N a 2 S  → H g S  +  2 N a C 2 H 3 O 2

First we have to find which is the limiting and which is the excess reagents.

This we do by calculating number  of moles

We see that for each 1 mole C 4 H 6 O 4 H g reacting, there is 1 mole of Na2s reacting and there is be formed 1 mole of HgS but 2 mole of 2 N a C 2 H 3 O 2

Step 2 : Calculating moles

mole = mass / Molar mass

⇒ mole C 4 H 6 O 4 H g = 34.23g / / mole =318.6780g 0.1074 mole

⇒ mole Na2S = 5.102g / 78.0452g/mole = 0.06537 mole

The limiting reactant is Na2S, which will be fully consumed after HgS and 2 N a C 2 H 3 O 2 is formed.

C 4 H 6 O 4 H g is the reactant in excess, after being consumed there will still be left.

⇒mole HgS : there will be produced 0.06537 mole HgS

⇒mole  N a C 2 H 3 O 2 : there will produced 0.13074 mole N a C 2 H 3 O 2

This means there will formed: 0.06537mole x 232.66g/mole = 15.2089842g  ≈15.21g HgS formed

Since there is consumed 0.06537 mole C 4 H 6 O 4 H g, there still remain: 0.1074- 0.06537 mole =0.04203 mole C 4 H 6 O 4 H g

mass C 4 H 6 O 4 H g = 0.04203 mole C 4 H 6 O 4 H g * 318.6780g/mole = 13.3940g ≈ 13.39g

So for C4H6O4Hg there remain 0.04203 mole in solution, which is 13.39g

Na2S is completely consumed, so there remain 0 moles

There got formed 0.06537 mole of HgS or 15.21g

and also 0.13074 mole of N a C 2 H 3 O 2 or  10.725g

43.452

step-by-step explanation:

284 times .153 = 43.452

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