The reactants are aluminum and iron nitrate.
The given solution of Mn²⁺ is 0.60 mg/mL.
Hence mass of Mn²⁺ in 5 mL of solution = 0.60 mg/mL x 5 mL = 3 mg
Molar mass of Mn = 54.9 g/mol
Hence, moles of Mn²⁺ = 3 x 10⁻³ g / 54.9 g/mol = 5.46 x 10⁻⁵ mol
The balanced equation for the reaction is,
2Mn²⁺ + 5KIO₄ + 3H₂O → 2MnO₄⁻ + 5KIO₃ + 6H⁺
The stoichiometric ratio between Mn²⁺ and KIO₄ is 2 : 5
Hence, moles of KIO₄ reacted = 5.46 x 10⁻⁵ mol x (5 / 2)
= 13.65 x 10⁻⁵ mol
Molar mass of KIO₄ = 230 g/mol
Hence needed mass of KIO₄ = 13.65 x 10⁻⁵ mol x 230 g/mol
= 0.031395 g
= 31.395 mg
≈ 31.4 mg
Answer:
The answer to your question is 7.4 moles of Aluminum
Explanation:
Data
moles of Al = ?
moles of Al₂O₃ = 3.7
Balanced chemical reaction
4 Al + 3 O₂ ⇒ 2 Al₂O₃
To solve this problem use proportions and cross multiplication. Use the coefficients of the balanced chemical equation.
4 moles of Aluminum ----------------- 2 moles of Al₂O₃
x ----------------- 3.7 moles of Al₂O₃
x = (3.7 x 4) / 2
x = 14.8 / 2
x = 7.4 moles of Aluminum
The answer could possibly be D.) The product have more potential than the activated complex OR C.) Forming the activated complex requires energy .
Elements that exceed the octet rule must have an unoccupied d orbital. I believe.
