In an experiment, 40.0 mL of 0.270M barium hydroxide was mixedwith 25.0mL of 0.3
ID: 684621 • Letter: I
Question
In an experiment, 40.0 mL of 0.270M barium hydroxide was mixedwith 25.0mL of 0.330 M aluminum sulfate. (a) Write the net ionic equation for the reaction that takesplace. (b) What is the total mass of precipitate that forms? (c) What are the molar concentrations of the ions that remainin the solution after the reaction is complete? In an experiment, 40.0 mL of 0.270M barium hydroxide was mixedwith 25.0mL of 0.330 M aluminum sulfate. (a) Write the net ionic equation for the reaction that takesplace. (b) What is the total mass of precipitate that forms? (c) What are the molar concentrations of the ions that remainin the solution after the reaction is complete?Explanation / Answer
a) The net ionic equation of the given reaction is - Ba+2 (aq) +SO4-2 (aq) --------> BaSO4(s) b) The number of moles of barium hydroxidepresent = 0.040L*0.270 mol/L = 0.0108 mol Similarly the number of moles of aluminum sulfate formed =0.025 L*0.330 mol/L = 0.00825 mol Therefore aluminum sulfate is limiting reactant. Therefore the number of moles of product would be formed.=0.00825 mol = 0.00825 mol * 233.32 g/mol = 1.924 g c) There are 0.0108 mol - 0.00825 mol = 0.00255mol barium hydroxide would be retained. Therefore concentration of the barium hydroxide = number ofmoles / Volume = 0.00255 mol /0.065 L = 0.0392 M
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