John the chemist is trying to make Mg3(PO4)2. He is doing this via the following unbalanced precipitation reaction.
K3PO4(aq) + MgCl2(aq)  Mg3(PO4)2(s) + KCl(aq)
John needs to make ¾ of a kg of Mg3(PO4)2. If he starts with 1500.mL of a 4.711 M solution of K3PO4 and 3500. mL of a solution with 325.72g of MgCl2 dissolved per Liter of solution. Will he be able to make at least ¾ of a kg of Mg3(PO4)2? How much of the product will precipitate out?

Question

John the chemist is trying to make Mg3(PO4)2.  He is doing this via the following unbalanced precipitation reaction.  
K3PO4(aq) + MgCl2(aq)  Mg3(PO4)2(s) + KCl(aq)
John needs to make ¾ of a kg of Mg3(PO4)2.  If he starts with 1500.mL of a 4.711 M solution of K3PO4 and 3500. mL of a solution with 325.72g of MgCl2 dissolved per Liter of solution.  Will he be able to make at least ¾ of a kg of Mg3(PO4)2? How much of the product will precipitate out?

anonymous · Answer

Here are the choices:

A. no, 467.2g         B. no, 380.0g         C.yes, 1.049kg         D. yes, 750.0g         E. yes, 928.6g

aaronq · Answer

moles K3PO4 = 4.711 M x 1.5 L = 7.066 moles

moles  MgCl2= [(325.72g)/(95.211 g/mol)]/1L x 3.5 L= 11.97 moles 

can you do the rest?

anonymous · Answer

Yes! Thank you =]

aaronq · Answer

(Y) no prob