Vol.250 before its to much pressure
- <u>The reaction that takes place is:</u>
Hg(NO₃)₂(ac) + Na₂S(ac) → HgS(s) + 2Na⁺ + 2NO₃⁻
Now we calculate the moles of each reagent -using the molecular weights-, in order to determine the limiting reactant:
- Moles of mercury (II) nitrate = 85.14 g *
=0.2622 moles. - Moles of sodium sulfide = 14.334 g *
=0.1837 moles.
Because the stoichiometric ratio between the reactants is 1:1, we compare the number of moles of each one upfront.
moles Hg(NO₃)₂ > moles Na₂S
<u>Thus Na₂S is the limiting reagent.</u>
So in order to find the mass of solid precipitate, we must calculate it using the moles of Na₂S:
![0.1837 molNa_{2} S*\frac{1molHgS }{1molNa_{2}S}*\frac{232.66g}{1molHgS} =42.740g](https://tex.z-dn.net/?f=0.1837%20molNa_%7B2%7D%20S%2A%5Cfrac%7B1molHgS%20%7D%7B1molNa_%7B2%7DS%7D%2A%5Cfrac%7B232.66g%7D%7B1molHgS%7D%20%3D42.740g)
The mass of the solid precipitate is 42.760 g.
- In order to calculate the grams of the reactant in excess that will remain after the reaction, we convert the moles that reacted into mass and substract them from the original mass:
Mass of Hg(NO₃)₂ remaining = ![85.14g-(0.1837molHg(NO_{3})_{2} * 324.7 g/mol)=25.49g](https://tex.z-dn.net/?f=85.14g-%280.1837molHg%28NO_%7B3%7D%29_%7B2%7D%20%2A%20324.7%20g%2Fmol%29%3D25.49g)
The mass of the remaning reactant in excess is 25.49 g.
- Because we assume complete precipitation, there are no more Hg⁺² or S⁻² ions in solution. The moles of NO₃⁻ and Na⁺ in solution remain the same during the reaction, so the number is calculated from the number added in the reactant:
Hg⁺²: 0 mol
NO₃⁻: ![0.2622molHg(NO_{3})_{2} *\frac{2molNO_{3}^{-}}{1molHg(NO_{3})_{2} *} =0.5244molNO_{3}^{-}](https://tex.z-dn.net/?f=0.2622molHg%28NO_%7B3%7D%29_%7B2%7D%20%2A%5Cfrac%7B2molNO_%7B3%7D%5E%7B-%7D%7D%7B1molHg%28NO_%7B3%7D%29_%7B2%7D%20%2A%7D%20%3D0.5244molNO_%7B3%7D%5E%7B-%7D)
Na⁺: ![0.1837molNa_{2} S*\frac{1molNa^{+}}{1molNa_{2}}=0.1837molNa^{+}](https://tex.z-dn.net/?f=0.1837molNa_%7B2%7D%20S%2A%5Cfrac%7B1molNa%5E%7B%2B%7D%7D%7B1molNa_%7B2%7D%7D%3D0.1837molNa%5E%7B%2B%7D)
S²⁻: 0 mol
Answer:
I hope that help you
Explanation:
energy is change from one form to another called _energy conversion
energy can neither be created nor destroyed it can only be transformed from one form to another called --The law of conservation of energy
cause to change in form character or function called - convert
the last sentensce
energy transformation
Required pH = 4.93
- OH⁻ from NaOH reacts with CH₃COOH giving CH₃COO⁻ and H₂O
- Let the volume of 3.5 M NaOH be x ml
Moles of NaOH = Moles of OH⁻ = Molarity * x ml = 3.5x mmol
- The reaction table for moles is as follows:
CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O
Initial 60 3.5x 40
Change -3.5x -3.5x +3.5x
Final (60-3.5x) 0 (40+3.5x)
- Substitute in Henderson equation and solve for x:
pH = pKa + log
![\frac{[CH_{3}COO^-]}{[CH_3COOH]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BCH_%7B3%7DCOO%5E-%5D%7D%7B%5BCH_3COOH%5D%7D%20)
4.93 = 4.76 + log
![\frac{(40+3.5x)}{(60-3.5x)}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%2840%2B3.5x%29%7D%7B%2860-3.5x%29%7D%20)
0.17 = log
![\frac{(40+3.5x)}{(60-3.5x)}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%2840%2B3.5x%29%7D%7B%2860-3.5x%29%7D%20)
![\frac{(40+3.5x)}{(60-3.5x)} = 1.479](https://tex.z-dn.net/?f=%20%5Cfrac%7B%2840%2B3.5x%29%7D%7B%2860-3.5x%29%7D%20%3D%201.479%20)
x = 5.62 ml NaOH required