Chemistry questions please! :)?

Question

# 89)

2K3PO4 (aq) + 3NiCl2 (aq) --> Ni3(PO4)2 (s) + 6KCl (aq)

What volume of 0.225 M K3PO4 solution is necessary to completely react with 134 mL of 0.0112 M NiCl2?

Sooo confused! Please explain so that I can understand in the future. Thanks. :)

Answer 1

2K3PO4 (aq) + 3NiCl2 (aq) --> Ni3(PO4)2 (s) + 6KCl (aq)

-1st you need to start with a known mol of a reagent

mol NiCl2 can be calculated by

mol NiCl2 = 0.134 * 0.0112 = 0.0015008 mol

-look at mol ratio in stoichiometric eqn. above.

mol NiCl2 : mol K3PO4 = 3:2

then mol K3PO4 that completely reacts = 0.0015008 * 2/3

= 0.001 mol

- Given that [K3PO4] is 0.225 M you can calculate the volume:

M stands for mol / L or mol / dm^3

therefore 0.225 M means that there is 0.225 mol in a 1L sol.

then 0.001 mol is in x L sol.

x = 0.001 / 0.225 = 0.044 L or 44 mL