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
Yes, refer to the previous answer.
Answer:
The answer to your question is Volume = 11.4 L
Explanation:
Data
Volume 1 = V1 = 6 L
Pressure 1 = P1 = 1 atm
Temperature 1 = T1 = 22°C
Volume 2 = V2 = ?
Pressure 2 = 0.45 atm
Temperature 2 = -21°C
Process
1.- Convert temperature (°C) to °K
T1 = 273 + 22 = 295°K
T2 = 273 + (-21) = 252°K
2.- Use the combined gas law to solve this problem
P1V1 / T1 = P2V2 / T2
-Solve for V2
V2 = P1V1T2 / T1P2
-Substitution
V2 = (6)(1)(252) / (295)(0.45)
- Simplification
V2 = 1512 / 132.75
- Result
V2 = 11.38 L
