1. KEY
GENERAL CHEMISTRY-I (1411)
S.I. # 6
1. Balance the following equations:
a. 2 CO(g) + __1__ O2 (g)  ___2__ CO2
b. __1 N2O5 (g) + __1__ H2O (l)  ___2__ HNO3 (aq)
c. __1 CH4 (g) + __4__ Cl2 (g)  ___1__ CCl4 (l) + ___4__ HCl (g)
d. __1 Al4C3 (s) + __12__ H2O (l)  ___4__ Al(OH)3 (s) + ___3__ CH4 (g)
e. __2 C5H10O2 (l) + __13__ O2 (g)  ___10__ CO2(g) + ___10__ H2O (l)
f. __2 Fe(OH)3 (s) + __3__ H2SO4(aq)  ___1__ Fe2(SO4)3 (aq) + __6_ H2O (l)
g. _1 Mg3N2 (s) + _4_H2SO4 (aq) _3__ MgSO4 (aq) + __1__ (NH4)2SO4 (aq)
2. What kind of reactions are the following?
a. HF + NaOH  H2O + NaF Acid-Base Neutralization
b. C9H5 + O2  H2O + CO2 Combustion
c. K+ + S2-  K2S Combination
d. CH3OH + O2  H2O + CO2 Combustion
e. H2O + NH3  HNO3 Combination
f. HCl + Mg  H2 + MgCl2 Acid+Active Metal
g. CuSO4  CuSO3 + O2 Decomposition
h. H2S + LiOH  Li2S + H2O Acid-Base Neutralization
i. H3PO4 + Ca  Ca3(PO4)2 + H2 Composition
3. Calculate the percent by mass of Oxygen in the following compounds:
a. SO3 FW=80.1 %O = [(3)(16amu) / (80.1 amu)]x100% = 59.9%
b. CH3COOCH3 FW=74 %O = [(2)(16amu) / (74 amu)]x100% = 43.2%
c. Cr(NO3)3 FW=238 %O = [(9)(16amu) / (238 amu)]x100% = 60.5%
d. sodium sulfate Na2SO4
FW=142.1 %O = [(4)(16amu) / (142.1 amu)]x100% = 45.0%
e. ammonium nitrate NH4NO3
FW=80 %O = [(3)(16amu) / (80 amu)]x100% = 60%

2. KEY
4. Calculate the following quantities:
a. mass, in grams, of 1.906x10-2 mol BaI2
FW=39.14g  (1.906x10-2 mol BaI2)(39.14g/1mol) = 7.455gBaI2
b. number of moles of NH4Cl in 48.3g
FW=53.49g  (48.3g NH4Cl)(1mol/53.49g) = 0.903gNH4Cl
c. number of molecules in 0.05752 mol HCHO2
(0.05752 mol HCHO2)(6.02214x1023 molecules/1mol) = 3.464x1022 HCHO2 molecules
d. number of O atoms in 4.88x10-3 mol Al(NO3)3
(4.88x10-3 mol Al(NO3)3)(9mol O /1mol Al(NO3)3)( 6.022x1023 atoms O/ 1mol O)
= 2.64x1022 O atoms
5. Determine the empirical formula of each of the following compounds if a sample
contains:
a. 0.104 mol K, 0.052 mol C, and 0.156 mol O
calculate the mole ratios
0.104 mol K / 0.052 = 2
0.052 mol C / 0.052 = 1
0.156 mol O / 0.052 = 3
The empirical formula is K2CO3
b. 5.28 g Sn and 3.37 g F
calculate the number of moles of each, and then the mole ratio
(5.28 g Sn)(1mol Sn / FW=118.7g) = 0.04448 mol Sn 0.04448/0.04448 = 1
(3.37 g F)(1mol F / FW=19g) = 0.1774 mol F 0.1774/0.04448 = 4
The empirical formula is SnF4
c. 87.5%N and 12.5%H by mass
Assume 100g sample, calculate moles then find mole ratio
87.5%N = (87.5 g N)(1mol N/14.01g N) = 6.25 mol N 6.25/6.25 = 1
12.5%H = (12.5 g H)(1mol H / 1.008 g H) = 12.4 mol H 12.4/6.25 = 2
The empirical formula is NH2