Respuesta :
Answer:
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
