SODIUM HYDROXIDE
IUPAC ID
Sodium hydroxide
Sodium oxidanide
Answer:
8 OH⁻(aq) + Mn(s) ⇒ MnO₄⁻(aq) + 4 H₂O(l) + 7 e⁻
Explanation:
Let's consider the following oxidation half-reaction that takes place in basic aqueous solution.
Mn(s) ⇒ MnO₄⁻(aq)
First, we will perform the mass balance. We will add 4 H₂O to the products side and 8 OH⁻ to the reactants side.
8 OH⁻(aq) + Mn(s) ⇒ MnO₄⁻(aq) + 4 H₂O(l)
Finally, we will perform the charge balance by adding 7 electrons to the products side.
8 OH⁻(aq) + Mn(s) ⇒ MnO₄⁻(aq) + 4 H₂O(l) + 7 e⁻
Hi there ,
Electron represents atomic number.
Protron and Neutron determine the weight.
Hope it helps.
Answer:
Mass = 58.96 g
Explanation:
Given data:
Mass of CH₄ = 21.5 g
Mass of O₂ = 387.5 g
Mass of CO₂ formed = ?
Solution:
Chemical equation:
CH₄ + 2O₂ → CO₂ + 2H₂O
Number of moles of CH₄:
Number of moles = mass / molar mass
Number of moles = 21.5 g/ 16 g/mol
Number of moles = 1.34 mol
Number of moles of O₂ :
Number of moles = mass / molar mass
Number of moles = 387.5 g/ 32 g/mol
Number of moles = 12.1 mol
now we will compare the moles of CO₂ with O₂ and CH₄.
O₂ : CO₂
2 : 1
12.1 : 1/2×12.1 = 6.05 mol
CH₄ : CO₂
1 : 1
1.34 : 1.34
Number of moles of CO₂ produced by CH₄ are less thus it will limiting reactant.
Mass of CO₂:
Mass = number of moles × molar mass
Mass = 1.34 mol × 44 g/mol
Mass = 58.96 g
Answer:
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Explanation: