Assumption University

Assumption University

Faculty of Engineering

Semester 2/2015

Quiz I

Course No. : BG 1108 General Chemistry

Lecturer : A. Kyi Kyi Tin

Section : 641

Date : 18 Feb, 2016

Time : (12:00-13:00)hrs

Name:………………………………………..Adm.Code:…………………Sec………………..

Calculators – Casio Fx-61 F, Sharp EL 556L/506R/506V and Texas Instrument TI-36X are permitted.

Answer all questions :

Part I : Short Questions 5 marks

Part II : Solve the Problems 25 marks

Total 30 marks

It is equivalent to 5% of the total marks for the course.

Attachment: Periodic Table

Part I: Answer the following short questions (3 questions: 5 marks)

1.  Explain the following briefly. (3 marks)
a. Isotopes: Same element having same atomic number but different mass number (or) same element having same number of protons and electrons but different number of neutrons.
b. Molecular Formula: An expression showing the exact numbers of atoms of each element in a molecule.
c. Excess Reagent: One or more reactants present in quantities greater than necessary to react with the quantity of the limiting reagent.
2.Give the number of electrons and write electron configuration of the following substances.
(1 mark each)
: e- = 24; 1s2,2s22p6,3s23p6,4s13d5
:e- = 36; 1s2,2s22p6,3s23p6,4s2,3d10,4p6
3. From the following molecular equation, write NET ionic equation and determine spectator ions also. (1 mark)
Ca(NO3)2(aq) + 2 KOH(aq) Ca(OH)2(s) + 2 KNO3(aq)
Ca2+ + 2NO3- + 2K+ +2OH- Ca(OH)2(s) + 2K+ +2NO3-
Ca2+ + 2OH- Ca(OH)2(s) (NET IONIC EQUATION)
spectator ions are: NO3- and K+

Part II: Solve the following problems (5 questions: 25 marks)

Instructions:

1.  Express your answers with 2 significant figures.

2.  R= 0.0821 L atm. K -1, mol -1

3.  STP = Standard Temperature 0˚C and Standard Pressure 760 mmHg.

No.1 (a) How many F atoms are in 8.54 g of F2?( 2 marks)
(b) The unknown compound with a molar mass of 155.06g/mol consists of
46.67% C, 7.80%H, and 45.72% Cl. Find the empirical formula and molecular formula for the compound. ( 3 marks)
(a) Molar mass of F2=(19*2) = 38g/mol
1 mol of F2= 6.02 *1023F2 molecules = 38g

(b) C : H : Cl
% by mass 46.67 : 7.80 : 45.72
by mole
Empirical formula [C3H6Cl]; Molecular formula =[C3H6Cl]x = 155.06 g/mol
[(12.01*3) +(1.008*6) +(35.5)]x=155.06
x = 2
Molecular formula = C6H12Cl2
No.2(a ) Given the reaction: SO3(g) + H2O(g) → H2SO4(l), which proceeds with a yield of 85.0%, calculate the amount of sulfuric acid(H2SO4) that is formed when 120. kg sulfur trioxide (SO3)is mixed with 32.4 kg water(H2O). (3 marks)
(b) Find the mass % of potassium in potassium permanganate (KMnO4)( 2 marks)
(a) SO3(g) + H2O(g) → H2SO4(l)
Equation (32+48)g + (2+16)g (2+32+64)g
80 g + 18 g 98 g
Problem 120 kg minimum amount
32.4 kg
Limit = SO3(g) ; Excess = H2O(g)

(b ) KMnO4 molar mass = (39+55+64) g/mol=158g/mol

No.3(a) A 35.5 mL 1.66 M HCl solution is mixed with 16.5 mL of 0.089M HCl solution. Calculate the concentration of the final solution.(3 marks)
(b) Calculate the density, in g/L of H2 gas at STP.(2 marks)
(a) total volume of the solution = (35.5+16.5)mL = 52mL
mole of 35.5 mL 1.66 M HCl =1.66 mol/L*35.5*10-3L=58.93*10-3 mol
mole of 16.5 mL 0.089 M HCl =0.089 mol/L*16.5*10-3L=1.47*10-3*mol
Total mole of the solution = (58.93+1.47)10-3mol=60.4*10-3mol

( b)

No.4(a) A sample of chlorine gas (Cl2)was collected over water at 21˚C and 680 mmHg. The volume of the container was 8.5 L. Calculate the mass of Chlorine collected. (Vapor pressure of water = 18 mmHg at 21˚C. (3 marks)
(b) How would you prepare 575 mL of MgSO4 (0.025M) from a stock solution of MgSO4 (0.045M) (2 marks)
(a) Cl2 gas T = (21+273)K=294K
P = (680-18)mmHg = (662/760)atm=0.87atm
V = 8.5L
mass of Cl2 (w) = ?
molar mass of Cl2 (M) = (35.5*2)=71g/mol

( b ) MgSO4 stock solution After Dilution
M1=0.045M M2 =0.025M
V1 = ? V2 = 575mL

No.5 Consider the reaction:
H2SO4 + 2 NaCl Na2SO4 + 2HCl
In one process 2.50 mol of H2SO4 and 328 g of NaCl are reacted.
Find: (a) Limiting agent
(b) % yield of the reaction if 165.8 g of HCl are actually produced.(Total 5 marks)
From Equation: H2SO4 : HCl From Equation: NaCl : HCl
Mole 1 : 2 Mole 2 : 2
From Problem 2.5 mole : ? 5.0 mole From Problem 328g(or) : ? 5.61 mole

(a) Limiting Agent = H2SO4 (The reactant which produces lesser quantity of product)
(b) Theoretical yield of HCl produced =
[Theoretical yield is the mass of product produced by the limiting agent]
% yield =

3