chem exercises X 2

1)
a)
NH4+ + OH- → NH3 + H­2O


b)
Consider the reaction of the excess NaOH with HCl:
NaOH + HCl → NaCl + H2O
No. of moles of HCl used = MV = 0.5 x (20/1000) = 0.01 mol
No. of moles of excess NaOH = 0.01 mol

Consider the reaction between the ammonium salt and NaOH.
NH4+ + OH- → NH3 + H­2O
Mole ratio OH- : NH3 = 1 : 1
No. of moles of OH- = No. of moles of NaOH
Total no. of moles of NaOH added = MV = 1.2 x (40/1000) = 0.048 mol
No. of moles of NaOH reacted = 0.048 - 0.01 = 0.038 mol
No. of moles of NH3 formed = 0.038 mol
Molar volume of NH3 at room temperature and pressure = 24 dm3 mol-1
Volume of NH3 formed = mol x (molar volume) = 0.038 x 24 = 0.912 dm3


c)
Molar mass of NH3 = 14 + 1x3 = 17 g mol-1
No. of moles of NH3 = 0.038 mol
Mass of NH3 = mol x (molar mass) = 0.038 x 17 = 0.646 g
Mass of the salt = 2.03 g
% by mass of NH3 in the salt = (0.646/2.03) x 100% = 31.8%

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2)
Molar mass of Cr = 52.0 g mol-1
Molar mass of Ag = 107.9 g mol-1
Molar mass of Cl = 35.5 g mol-1

Fraction by mass of Cl in AgCl = 35.5/(107.9+35.5) = 35.5/143.4
Mass of Cl in 0.3383 g of AgCl = 0.3383 x (35.5/143.4) = 0.08375 g

In 0.1246 g of the compound:
Mass of Cl = 0.08375 g
Mass of Cr = 0.1246 - 0.08735 = 0.04085
Mole ratio Cr : Cl = 0.04085/52 : 0.08375/35.5 = 1 : 3

Ans: Empirical formula of the compound = CrCl3
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