Can someone me with this. i am lost and i don't understand how can i do this.
what volume, in l, of 0.219 m nicl2 solution is required to produce 68.4 g of precipitate.

3nicl2(aq) + 2na3po4(aq) → ni3(po4)2(s) + +6nacl(aq)

2.56 L NiCl2

Explanation:

This is a stoichiometry problem. You need to find a way to go from g of Ni3(PO4)2 to L of NiCl2. To do this you travel

mass Ni3(PO4)2 > moles Ni3(PO4)2 > moles NiCl2 > L NiCl2

so you need to know the key that lets you travel from one to the next.

Mass > moles, use molar mass

moles > moles, use the mole ratio

moles > L, use the molarity (moles / liter) in this case. (With gases, you use the STP gas volume of 22.4L / mole).

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