1 mols of Aluminium ion forms 1 mol aluminium phosphate
Molar mass of AlPO_4
Moles of AlPO_4
- 61µg/106
- 0.000061/106
- 5.75×10^{-7}
- 57.5µmol
Moles of Al3+=57.5µmol
Answer:
The answer to your question is 8.74 g of He
Explanation:
Data
V = 2.4 x 10² L
P = 99 kPa
T = 0°C
mass = ?
Process
1.- Convert kPa to atm
P = 99 kPa = 99000 Pa
1 atm --------------- 101325 Pa
x --------------- 99000 Pa
x = (99000 x 1) / 101325
x = 0.977 atm
2.- Convert temperature to °K
°K = 273 + 0
°K = 273
3.- Substitution
PV = nRT
- Solve for n
n = PV / RT
n = (0.977)(2.4 x 10²) / (0.082)(273)
n = 24.48 / 22.386
n = 1.093 moles
4.- Calculate the grams of He
8 g -------------------- 1 mol
x -------------------- 1.093 moles
x = (1.093 x 8) / 1
x = 8.74 g
Answer: Li is the reducing agentg and O is the oxidizing agent.
Explanation:
1) The oxidizing agent is the one that is reduced and the reducing agent is the one that is oxidized.
2) The given reaction is:
4Li(s) + O₂ (g) → 2 Li₂O(s)
3) Determine the oxidation states of each atom:
Li(s): oxidation state = 0 (since it is alone)
O₂ (g): oxidation state = 0 (since it is alone)
Li in Li₂O (s) +1
O in Li₂O -2
That because 2× (+1) - 2 = 0.
4) Determine the changes:
Li went from 0 to + 1, therefore it got oxidized and it is the reducing agent.
O went from 0 to - 2, therefore it got reduced and it is the oxidizing agent.
