Can anyone help?

[匿名]

2018-04-04 11:39 am

Near to impossible

更新1:

0.1938-g of a chlorocarbon compound is analyzed by burning it in oxygen,collecting the evolved gases in a solution of NaOH. After neutralizing, its was treated w. 31.30 mL of a 0.1623 M AgNO3 soln. This precipitated the chloride (Cl-) out as AgCl and left an excess of AgNO3. The excess AgNO3 is titrated w. 0.1260M KSCN & required 21.22 mL to reach endpoint in a Volhard titration. Calculate the % w/w Cl– (35.45 g/mol) in sample. Cl– + Ag+ → AgCl(s) Ag+ + SCN– → AgSCN(s)

更新2:

Chem help plz

回答 (1)

wanszeto

2018-04-04 12:23 pm

The sequence of reactions :
[1] Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
[2] Ag⁺(aq) + SCN⁻(aq) → AgSCN(s)
(Ag⁺ ions are consumed in both reactions [1] and [2].)

Consider reaction [2]:
Mole ratio Ag⁺ : SCN⁻ = 1 : 1
No. of moles of SCN⁻ reacted = (0.1260 mol/L) × (21.22/1000 L) = 0.0026737 mol
No. of moles of Ag⁺ consumed in reaction [2] = 0.0026737 mol

Consider reaction [1]:
Mole ratio Ag⁺ : Cl⁻ = 1 : 1
Total number of moles of Ag⁺ added = (0.1623 mol/L) × (31.30/1000 L) = 0.0050800 mol
Number of moles of Ag⁺ consumed in reaction [1] = (0.0050800 - 0.026737) mol = 0.0024063mol
Number of moles of Cl⁻ reacted = 0.0024063 mol

In 0.1938 g of the chlorocarbon:
Number of moles of Cl = 0.00240627 mol
Mass of Cl = (0.0024063 mol) × (35.45 g/mol) = 0.08530 g
% w/w Cl = (0.08530/0.1938) × 100% = 44.01%

👍 3 👎 0