Will Give Badge and Fan !

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?

Question

Will Give Badge and Fan !

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?

anonymous · Answer

The solution for problem 3 was that the concentration of Lithium was 0.1632 Molars and the amount of solution was 200 mL

anonymous · Answer

@adrynicoleb @BugzytheGreat @ganeshie8 @thomaster

micahm · Answer

what the question

micahm · Answer

do you have to explain it if so please put this in your own words

anonymous · Answer

We have to find " grams of the new solid precipitates?"

We know Lithium's Molar concentration in Li3PO4 = 0.1632 M

The total amount of solution was 200 mL = 0.2 L . 

Next , the net ionic equation was : 3Mg+(aq) + 2PO4(aq) -> Mg3(PO4)2

We have to find the grams of Mg3(PO4)2 that form

anonymous · Answer

@adrynicoleb @HelpBlahBlahBlah @ganeshie8 @thomaster

micahm · Answer

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anonymous · Answer

Uh?

micahm · Answer

that should help you

anonymous · Answer

I can't understand that . Can you paragraph it or give me the link

micahm · Answer

https://www.google.com/search?q=An+aqueous+solution+containing+excess+magnesium+chloride+is+mixed+with+the+saturated+solution+of+lithium+phosphate+prepared+in+problem+%233.+In+problem+%234%2C+you+wrote+the+net+ionic+equation+for+the+reaction+that+results.+How+many+grams+of+the+new+solid+precipitates%3F&rlz=1C1GGGE___US571US573&oq=An+aqueous+solution+containing+excess+magnesium+chloride+is+mixed+with+the+saturated+solution+of+lithium+phosphate+prepared+in+problem+%233.+In+problem+%234%2C+you+wrote+the+net+ionic+equation+for+the+reaction+that+results.+How+many+grams+of+the+new+solid+precipitates%3F&aqs=chrome..69i57&sourceid=chrome&es_sm=0&ie=UTF-8

micahm · Answer

click it

anonymous · Answer

@Destinycholie @Cuanchi

destinycholie · Answer

Use the Law of Conservation of Mass: 
 (50.00 g total - 46.13 g C) = 3.87 g H 
 (3.87 g H) / (1.007947 g H/mol) = 3.84 mol H

anonymous · Answer

But how can we apply it here : 

We have to find " grams of the new solid precipitates?"

We know Lithium's Molar concentration in Li3PO4 = 0.1632 M

The total amount of solution was 200 mL = 0.2 L . 

Next , the net ionic equation was : 3Mg+(aq) + 2PO4(aq) -> Mg3(PO4)2

We have to find the grams of Mg3(PO4)2 that form

anonymous · Answer

@Cuanchi

destinycholie · Answer

http://www.google.com/url?sa=t&rct=j&q=&esrc=s&frm=1&source=web&cd=3&ved=0CCoQFjAC&url=http%3A%2F%2Fcollege.hmco.com%2Finstructors%2Fcatalog%2Fwalkthroughs%2Fpdf%2F0618447962_ch04.pdf&ei=wllKVPmZAveZsQSIoYCADg&usg=AFQjCNGO4xCtttwyEIhMpcZrWJP9bpJwgg&sig2=7W1UkFM4_fppqjUMou3BtA

destinycholie · Answer

see if that helps

cuanchi · Answer

because you have an excess of Mg Cl, your limiting reactant is the Li3PO4 in your solution.
1. Calculate the moles "n" of Li3PO4 in the 0.2L of 0.1632M solution by multiplying molarity by volume in liters. 
2. then according to the balanced reaction you will have a 2 moles of Li3PO4 will form one mole of Mg3(PO4)2; then you divide the "n" moles by 2 and that is going to be the moles of Mg3(PO4)2 precipitated.
3. multiply the moles of Mg3(PO4)2 by the molecular mass of Mg3(PO4)2 and you will get the grams of Mg3(PO4)2 formed.