CHEM - VOLUMETRIC ANALYSIS

Scarbo

2011-07-31 1:08 am

1) A powder mixture of sodiumhydroxide and anhydrous sodium carbonate required 75.0 cm3 of 0.200M sulphuric acid for complete reaction. The volume of carbon dioxide evolved (correctedto s.t.p.) was 112cm3. Calculate the composition by mass of theoriginal mixture. ANS: 0.80g NaOH, 0.53gNa2Co3 2) 100cm3 of a solutionB containing nitric acid and hydrochloric acid requires 68.0cm3 of 1.00M sodiumhydroxide solution for exact neutralization. 50.0cm3 of b reactedwith excess sliver nitrate solution to give 2.725g of sliver chlorideprecipitate. Find the concentration (in g dm-3) of each acid in B. ANS: HCl: 13.9g dm-3,HNO3: 18.9g dm-3 3) potassium permangate solutionoxidizes iron(II) salt to iron(III) salt according to the following ionicequation: 5Fe2+ (aq)+ MnO4-(aq)+8H+(aq)à 5Fe3+(aq)+ Mn2+(aq)+H2O(l)1.817g of potassium permanganate are dissolved in water to give250.0 cm3 of aqueous solution. 2.185 g of an iron(II) salt are dissolved inwater and the solution made up to 100cm3. 40.00 cm3 of this iron(II) saltsolution require 25.00cm3 of the permanganate solution for complete reaction. a) what is the molarity of the potassium permanganate solution? ANS: 0.04600Mb) What is the molarity of the Fe2+ ions in the iron(II) salt solution? ANS: 0.1438Mc) Given that 1 mole of the iron(II) salt contains 1 mole of Fe2+ions, what is the molar mass of theiron (II) salt? ANS: 152.0gd) What is the percentage by mass of iron in the salt? ANS: 36.84% HOW CAN I FIND THE ANSWER? PLEASE GIVESOLUTION. THANKS!!

回答 (1)

不用客氣

2011-07-31 6:55 am

✔ 最佳答案

1)
Na2CO3 + H2SO4 → Na2SO4+ H2O + CO2
No. of moles of CO2 formed = 112/22400 = 0.005 mol
No. of moles of H2SO4 reacted with Na2CO3= 0.005 mol
No. of moles of Na2CO3 = 0.005 mol
Mass of Na2CO3 = 0.005 x (23x2 + 12 + 16x3) = 0.53 g

2NaOH + H2SO4 → Na2SO4+ 2H2O
Total no. of moles of H2SO4 = 0.2 x (75/1000) = 0.015 mol
No. of moles of H2SO4 reacted with Na2CO3= 0.005 mol
No. of moles of H2SO4 reacted with NaOH = 0.015 - 0.005 =0.01 mol
No. of moles of NaOH = 0.01 x 2 = 0.02 mol
Mass of NaOH = 0.02 x (23 + 16 + 1) = 0.80g

=====
2)
HCl + AgNO3 → AgCl + HNO3
No. of moles of AgCl formed = 2.725/(107.9 + 35.5) = 0.019 mol
No. of moles of HCl reacted with HCl = 0.019 mol
Concentration of HCl = [0.019 x (1 + 35.5)] / (50/1000) = 13.9 g dm⁻³

HCl + NaOH → NaCl + H2O
No. of moles of HCl in 100 cm³ of the solution = 0.019 x (100/50) = 0.038 mol
No. of moles of NaOH reacted with HCl = 0.038 mol

HNO3 + NaOH → NaNO3 + H2O
Total no. of moles of NaOH used = 1 x (68/1000) = 0.068 mol
No. of moles of NaOH reacted with HCl = 0.038 mol
No. of moles of NaOH reacted with HNO3 = 0.068 - 0.038 = 0.03 mol
Concentration of HNO3 = [0.03 x (1 + 14 + 16x3)] / (100/1000) = 18.9 g dm⁻³

=====
3)
a)
Molarity of KMnO4 = [1.817/(39 + 55 + 16x4)] / (250/1000) = 0.04600 M

b)
5Fe^2+(aq) +MnO4^-(aq) + 8H^+(aq) → 5Fe^3+(aq) +Mn^2+(aq) + H2O(l)
No. of moles of KMnO4 = 0.046 x (25/1000) = 0.00115 mol
No. of moles of MnO4^- ions = 0.00115 mol
No. of moles of Fe^2+ ions = 0.00115 x 5 = 0.00575 mol
Molarity of Fe^2+ ions = 0.00575 / (40/1000) = 0.1438 M

c)
No. of moles of Fe^2+ in 2.185 g of iron(II) salt = 0.00575 x (100/40) = 0.014375mol
No. of moles of 2.185 g of iron(II) salt = 0.014375 mol
Molar mass of the iron(II) salt = 2.185/0.014375 = 152.0 g

d)
No. of moles of Fe^2+ in 2.185 g of iron(II) salt = 0.014375 mol
Mass of Fe in 2.185 g of iron(II) salt = 0.014375 x 56 = 0.805 g
% by mass of Fe in iron(II) salt = (0.805/2.185) x 100% = 36.84%

參考: micatkie

👍 0 👎 0