Respuesta :
The number of solid precipitate that will be formed is 37.08 g
calculation
write the equation for reaction
=Hg(NO3)2 +Na2SO4 = HgSO4(s) +2NaNO3(aq)
find the moles of each reactant
moles ofHg(NO3)2=126.27/324.6= 0.389 moles
moles of Na2SO4=17.796/142=0.125 moles
NaSO4 is the limiting reagent and by use of mole ratio of NaSO4:HgSO4 which is 1:1 therefore the moles of H2SO4 is also= 0.125 moles
mass HgSO4=moles x molar mass
=0.125 x296.65= 37.08g
Answer: The mass of mercury (II) sulfide is 53.51 g
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
- For mercury (II) nitrate:
Given mass of mercury (II) nitrate = 126.27 g
Molar mass of mercury (II) nitrate = 324.7 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of mercury (II) nitrate}=\frac{126.27g}{324.7g/mol}=0.388mol[/tex]
- For sodium sulfide:
Given mass of sodium sulfide = 17.796 g
Molar mass of sodium sulfide = 78.04 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of sodium sulfide}=\frac{17.796g}{78.04g/mol}=0.23mol[/tex]
The chemical equation for the reaction of Mercury (II) nitrate and sodium sulfide follows:
[tex]Hg(NO_3)_2(aq.)+Na_2S(aq.)\rightarrow HgS(s)+2NaNO_3(aq.)[/tex]
By Stoichiometry of the reaction:
1 mole of sodium sulfide reacts with 1 mole of mercury (II) nitrate
So, 0.23 moles of sodium sulfide will react with = [tex]\frac{1}{1}\times 0.23=0.23mol[/tex] of mercury (II) nitrate
As, given amount of mercury (II) nitrate is more than the required amount. So, it is considered as an excess reagent.
Thus, sodium sulfide is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 mole of sodium sulfide produces 1 mole of mercury (II) sulfide
So, 0.23 moles of sodium sulfide will produce = [tex]\frac{1}{1}\times 0.23=0.23moles[/tex] of mercury (II) sulfide
Now, calculating the mass of mercury (II) sulfide from equation 1, we get:
Molar mass of mercury (II) sulfide = 232.66 g/mol
Moles of mercury (II) sulfide = 0.23 moles
Putting values in equation 1, we get:
[tex]0.23mol=\frac{\text{Mass of mercury (II) sulfide}}{232.66g/mol}\\\\\text{Mass of mercury (II) sulfide}=53.51g[/tex]
Hence, the mass of mercury (II) sulfide is 53.51 g
