- #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

Note - The molarity of the prepared lithium phosphate mentioned above from a previous problem was

So from the reaction equation -

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 =

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

**(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.**__after the exam__

My solution**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?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

__(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 ..__**0.163 moles/litre****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 =

__(to 3 significant figures)__**44.0 g**Is this correct ? If not where have I gone wrong ? Can anybody help ?

Regards,

Jack