Answer:
1. KCLO3------>KCL + 3/2O2(g)
2. 122.5g/mol
3. 0.2mol
4. 18.5g
Mass of Co(NO₃)₂ = 1.95 g
V KOH = 0.350 L
[KOH] = 0.220 M
Kf = 5.0 x 10⁹
molar mass of Co(NO₃)₂ = 182.943 g/mol
so [Co(NO₃)₂] = 1.95 / (0.350 * 182.943) = 0.03045 M
[Co²⁺] = 0.03045 M
[OH⁻] = 0.22 M
chemical reaction:
Co²⁺(aq) + 4 OH⁻ ⇄ Co(OH)₄²⁻
I (M) 0.03045 0.22 0
C (M) - 0.03045 - 4 (0.03045) 0.03045
E (M) - x 0.22 - 4(0.03045) 0.03045
= 0.0982
Kf = [Co(OH)₄²⁻] / [Co⁺²][OH⁻]⁴
5.0 x 10⁹ = (0.03045) / x (0.0982)⁴
x = 6.5489 x 10⁻⁸
at equilibrium:
[Co²⁺] = 6.54 x 10⁻⁸
[OH⁻] = 0.0982 M
[Co(OH)₄²⁻] = 0.03045 M
There is 3800 mmHg in 5 atm
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