CuSO4(aq) + Na2CO3(aq) → CuCO3(s) + Na2SO4(aq) How many mL of 0.120 M Na2CO3 would be needed to precipitate all of the copper ions in 25.7 mL of 0.149 M CuSO4? How many grams of CuCO3 could be recovered?
... How many mL of 0.120 M Na2CO3 would be needed to precipitate all of the copper ions in 25.7 mL of 0.149 M CuSO4? ... 1. The relationship M1V1 = M2V2 will be used 2. M1 = 0.120 M Na2CO3 V1 = unknown 3. M2 = 0.149 M CuSO4 V2 = 25.7 mL 4. Rearrange the equation in 1 to solve for V1: V1 = (M2V2)/M1 5. Plug in the numbers and calculate V1: V1 = ((0.149 M CuSO4)*(25.7 mL of CuSO4)/(0.120 M Na2CO3) V1 = 31.9 mL of Na2CO3 ... How many grams of CuCO3 could be recovered? ... This will be the weight equivalent to the number of moles precipitated. ... all of the copper ions in 25.7 mL of 0.149 M CuSO4 ... is used to calculate the moles of copper salt produced (the number of moles of copper will be the same in the reactant and product). moles of copper in CuSO4 = ((25.7 mL) * (1 L/1000 mL)) * (0.149 M) = 0.00383 moles The same number of moles of CuCO3 can be precipitated. Formula weight of CuCO3 = 123.6 g/mole (to four sig figs) (63.546 + 12.011 + 3*15.9994 = 123.5552 ) Weight of CuCO3 that can be produced = ( 0.00383 moles) * (123.6 g/mole) = 0.473 g
The carbonate ion seams an appealing alternative, but how is this carried out commercially - seawater is a well recognized source of Mg.The Mg2+ cation is the second most considerable cation in seawater (occurring at about 12% of the mass of sodium there), which makes seawater and sea-salt an appealing industrial supply of Mg. To extract the magnesium, calcium hydroxide is brought to seawater to kind magnesium hydroxide precipitate. MgCl2 + Ca(OH)2 → Mg(OH)2 + CaCl2 Magnesium hydroxide (brucite) is insoluble in water so it can be filtered out, and reacted with hydrochloric acid to acquire centred magnesium chloride. Mg(OH)2 + 2 HCl → MgCl2 + 2 H2O From magnesium chloride, electrolysis produces magnesium. So I offer as an answer: hydroxide ion.
