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).

The attached picture shows what this problem looks like.

we need the questions , i may then you.

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml