Asked by LARA
Determine the concentration of a solution of Mg(NO3)2 given the precipitate formed is 2.97 g and there are 50mL of Mg(NO3)2 and 50 mL of NaOH. Please show all work for each step clearly
Answers
Answered by
DrBob222
Mg(NO3)2 + 2NaOH ==> Mg(OH)2 + 2NaNO3
mols Mg(OH)2 formed = 2.97/molar mass Mg(OH)2.
1 mol Mg(OH)2 was formed from 1 molo Mg(NO3)2.
M Mg(NO3)2 = mols Mg(NO3)2/L soln
Note: 50 mL each soln = 100 mL total or 0.1 L.
This calculation is based upon Mg(NO3)2 and NaOH being in stoichiometric proportions; i.e. no common ion effect due to an excess of either.
mols Mg(OH)2 formed = 2.97/molar mass Mg(OH)2.
1 mol Mg(OH)2 was formed from 1 molo Mg(NO3)2.
M Mg(NO3)2 = mols Mg(NO3)2/L soln
Note: 50 mL each soln = 100 mL total or 0.1 L.
This calculation is based upon Mg(NO3)2 and NaOH being in stoichiometric proportions; i.e. no common ion effect due to an excess of either.
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