Alum Base
HELP ! College Chemistry !?
1. Net Ionic Equation- What numbers replace the question marks ?
0Cs+(aq) + ?I -(aq) + 1Cl2(aq) = 0Cs+(aq) + ?Cl -(aq) + 1I2(aq)
2. If 26.13 mL of a standard 0.1650 M NaOH solution is required to neutralize 33.47 mL of H2SO4, what is the molarity of the acid solution?
Answers: .064 & .1288 are both wrong. So, how exactly do I do this problem ?
3. Aluminum sulfate, known as cake alum, has a remarkably wide range of uses, from dyeing leather and cloth to purifying sewage. In aqueous solution, it reacts with base to form a white precipitate.
Overall Equation- 1Al2(SO4)3+6NaOH=3Na2SO4+2Al(OH)3
Net Ionic- 1[Al]3+ + 3[OH]- = 1Al(OH)3
What mass of precipitate forms when 119.5 mL of 0.680 M NaOH is added to 517 mL of a solution that contains 16.5 g aluminum sulfate per liter?
Answers- 19.01 & 5.77 are the wrong answers. So, how do I do this problem ?
1. Net Ionic Equation- What numbers replace the question marks ?
0Cs+(aq) + 2 I -(aq) + Cl2(aq) = 0Cs+(aq) + 2 Cl -(aq) + I2(aq)
2. equation 2NaOH + H2SO4 –> Na2SO4 + 2 H2O
So each mole of H2SO4 is equivalent to 2 moles of NaOH
Moles of NaOH = 26.13/1000 L x 0.1650 = 0.0043 Moles.
Moles of H2SO4 = 0.0043 / 2 = 0.00215
But this also equals 33.47/1000 L x M
M = 1000 x 0.00215 / 33.47 = 0.0642 M (You need 4 sig figs)
3. From the stoichiometry 6 NaOH are equivalent to 1 Al2(SO4)3
Moles of NaOH = (119.5/1000) x 0.680 = 0.08126 moles
Moles of Al2(SO4)3 = (517/1000 ) x 16.5 / 342.2 = 0.025 moles.
The Al2(SO4)3 is in excess as 0.025 moles needs 6 x 0.025 moles of NaOH to fully react and there is less NaOH than that. NaOH is the limiting reactant.
From the equation 3 NaOH produce Al(OH)3
so 0.08126 molesof NaOH —> 0.02709 moles Al(OH)3 = 0.02709 x 78 = 2.113g Al(OH)3
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