Answer:
The answer to your question is: letter D. 1.33 L
Explanation:
Data
V1 = 50 ml
C1 = 19.3
To solve this problem use the formula C₁V₁ = C₂V₂
C2 = C1V1 / V2
C = concentration
V = volume
a) 1.15 L
C2 = (19.3)(50) / 1150
C2 = 0.84 M
b) No right answer
c) V2= 0.80 L
C2 = (19.3)(50) / 800
C2 = 1.2 M
d) V2 = 1.33 L
C2 = (19.3)(50) / 1330
C2 = 0.72 M
e) V2 = 350 ml
C2 = (19.3)(50) / 350
C2 = 2.75 M
The balanced equation for the above reaction is
2K₃PO₄ + 3NiCl₂ ---> 6KCl + Ni₃(PO₄)₂
stoichiometry of K₃PO₄ to NiCl₂ is 2:3
the number of NiCl₂ moles reacted - 0.0110 mol/L x 0.154 L = 1.69 x 10⁻³ mol
if 3 mol of NiCl₂ reacts with - 2 mol of K₃PO₄
then 1.69 x 10⁻³ mol of NiCl₂ reacts with - 2/3 x 1.69 x 10⁻³ = 1.13 x 10⁻³ mol of K₃PO₄
molarity of K₃PO₄ solution given - 0.205 M
there are 0.205 mol in 1 L
therefore 1.13 x 10⁻³ mol are in - 1.13 x 10⁻³ mol / 0.205 mol/L = 5.51 mL
volume of K₃PO₄ required - 5.51 mL
