What is the mass in grams of a 8.4 mole sample of iron?
2 answers:
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Answer:
A) 0.065 M is its molarity after a reaction time of 19.0 hour.
B) In 52 hours will react 69% of its initial concentration.
Explanation:
The reaction is first order in :
Initial concentration of =
a) Final concentration of after 19.0 hours=
t = 19.0 hour = 19.0 × 3600 seconds ( 1 hour = 3600 seconds)
Rate constant of the reaction = k =
The integrated law of first order kinetic is given as:
0.065 M is its molarity after a reaction time of 19.0 h.
b)
Initial concentration of =
Final concentration of after t =
Rate constant of the reaction = k =
The integrated law of first order kinetic is given as:
t = 185,902.06 s = ≈ 52 hours
In 52 hours will react 69% of its initial concentration.
First a balanced reaction equation must be established:
→ 
Now if mass of aluminum = 145 g
the moles of aluminum = (MASS) ÷ (MOLAR MASS) = 145 g ÷ 30 g/mol
= 4.83 mols
Now the mole ratio of Al : O₂ based on the equation is 4 : 3
[4Al + 3 O₂ → 2 Al₂O₃]
∴ if moles of Al = 4.83 moles
then moles of O₂ = (4.83 mol ÷ 4) × 3
= 3.63 mol (to 2 sig. fig.)
Thus it can be concluded that 3.63 moles of oxygen is needed to react completely with 145 g of aluminum.
Now if mass of aluminum = 145 g
the moles of aluminum = (MASS) ÷ (MOLAR MASS) = 145 g ÷ 30 g/mol
= 4.83 mols
Now the mole ratio of Al : O₂ based on the equation is 4 : 3
[4Al + 3 O₂ → 2 Al₂O₃]
∴ if moles of Al = 4.83 moles
then moles of O₂ = (4.83 mol ÷ 4) × 3
= 3.63 mol (to 2 sig. fig.)
Thus it can be concluded that 3.63 moles of oxygen is needed to react completely with 145 g of aluminum.