- #1
jackthehat
- 41
- 5
- Homework Statement
- To calculate the mass of a precipitate where one of the solutes has been previously prepared
- Relevant Equations
- 3MG 2+ + 2PO4 3- ---> Mg3(PO4)2 (Net Ionic equation)
MgCl2 + Li3PO4 ---> Mg3(PO4)2 + LiCl (Precipitation reaction equation)
Hi there,
I have a problem from an exam which I passed recently, however this was one problem I got wrong and I want to understand how to solve the thing. Now I shall detail the problem below. This is my subsequent attempt to solve it after the exam (so I don't know if this attempt is correct or not) and would appreciate any input reassuring me that this, my last attempt, is correct or if wrong where I am going wrong with this type of problem.
Problem
An aqueous solution containing excess magnesium chloride is mixed with the saturated solution of lithium phosphate prepared in problem #3. In problem #4, you wrote the net ionic equation for the reaction that results. How many grams of the new solid precipitates?
My solution
Note - The molarity of the prepared lithium phosphate mentioned above from a previous problem was 0.163 moles/litre (I know this is correct from the exam feedback). Also the net ionic equation talked about in the problem is listed in the equations section above, but I repeat here ..
3MG 2+ + 2PO4 3- ---> Mg3(PO4)2
So from the reaction equation - MgCl2 + Li3PO4 ---> Mg3(PO4)2 + LiCl
we can see that .. 1 mole of LI3PO4 produces 1 mole of Mg3(PO4)2 (which is the precipitate) .. so the molar ratio is 1/1 = 1
Now the previously prepared saturated solution of Li3PO4 has a calculated molarity of 0.163 M so as the molar ratio of this to our saturate is 1:1
then the molarity of Mg3(PO4)2 is also 0.163 M.
So we now convert from moles to mass (grams) of the precipitate by ..
Calculating it's molar mass (g/mol) and multiplying this by it's actual molarity (mol) giving the mass present of precipitate in solution after the reaction.
Molar mass of Mg3(PO4)2 = ((3 x 24.305) + (2 x 30.974) + (8 x 15.999)) g/mol = 269.855 g/mol
Multiplying this now by molarity we get ... (269.855) g/mol x (0.163) mol = 43.986 = 44.0 g (to 3 significant figures)
Is this correct ? If not where have I gone wrong ? Can anybody help ?
Regards,
Jack
I have a problem from an exam which I passed recently, however this was one problem I got wrong and I want to understand how to solve the thing. Now I shall detail the problem below. This is my subsequent attempt to solve it after the exam (so I don't know if this attempt is correct or not) and would appreciate any input reassuring me that this, my last attempt, is correct or if wrong where I am going wrong with this type of problem.
Problem
An aqueous solution containing excess magnesium chloride is mixed with the saturated solution of lithium phosphate prepared in problem #3. In problem #4, you wrote the net ionic equation for the reaction that results. How many grams of the new solid precipitates?
My solution
Note - The molarity of the prepared lithium phosphate mentioned above from a previous problem was 0.163 moles/litre (I know this is correct from the exam feedback). Also the net ionic equation talked about in the problem is listed in the equations section above, but I repeat here ..
3MG 2+ + 2PO4 3- ---> Mg3(PO4)2
So from the reaction equation - MgCl2 + Li3PO4 ---> Mg3(PO4)2 + LiCl
we can see that .. 1 mole of LI3PO4 produces 1 mole of Mg3(PO4)2 (which is the precipitate) .. so the molar ratio is 1/1 = 1
Now the previously prepared saturated solution of Li3PO4 has a calculated molarity of 0.163 M so as the molar ratio of this to our saturate is 1:1
then the molarity of Mg3(PO4)2 is also 0.163 M.
So we now convert from moles to mass (grams) of the precipitate by ..
Calculating it's molar mass (g/mol) and multiplying this by it's actual molarity (mol) giving the mass present of precipitate in solution after the reaction.
Molar mass of Mg3(PO4)2 = ((3 x 24.305) + (2 x 30.974) + (8 x 15.999)) g/mol = 269.855 g/mol
Multiplying this now by molarity we get ... (269.855) g/mol x (0.163) mol = 43.986 = 44.0 g (to 3 significant figures)
Is this correct ? If not where have I gone wrong ? Can anybody help ?
Regards,
Jack