Balance the following redox reaction if it occurs in acidic solution. what are the coefficients in front of h+ and fe3+ in the balanced reaction? fe2+(aq) + mno4-(aq) → fe3+(aq) + mn2+(aq) balance the following redox reaction if it occurs in acidic solution. what are the coefficients in front of h+ and fe3+ in the balanced reaction? fe2+(aq) + mno4-(aq) → fe3+(aq) + mn2+(aq) h+ = 8, fe3+ = 1 h+ = 8, fe3+ = 5 h+ = 2, fe3+ = 3 h+ = 3, fe3+ = 2 h+ = 5, fe3+ = 1

H⁺ = 8; Fe³⁺ = 5

Explanation:

In order to balance a redox reaction, we use the ion-electron method.

Step 1: Identify both half-reactions.

Reduction: MnO₄⁻(aq) → Mn²⁺(aq)

Oxidation: Fe²⁺(aq) → Fe³⁺(aq)

Step 2: Balance the masses using H⁺ and H₂O where appropiate.

8 H⁺(aq) + MnO₄⁻(aq) → Mn²⁺(aq) + 4 H₂O(l)

Fe²⁺(aq) → Fe³⁺(aq)

Step 3: Balance the charges adding electrons where appropiate.

8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

Fe²⁺(aq) → Fe³⁺(aq) + 1 e⁻

Step 4: Multiply both half-reactions by numbers that assure that the number of electrons gained is equal to the number of electrons lost.

1 × [8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)]

5 × [Fe²⁺(aq) → Fe³⁺(aq) + 1 e⁻]

Step 5: Add both half-reactions

8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ + 5 Fe²⁺(aq) → Mn²⁺(aq) + 4 H₂O(l) + 5 Fe³⁺(aq) + 5 e⁻

8 H⁺(aq) + MnO₄⁻(aq) + 5 Fe²⁺(aq) → Mn²⁺(aq) + 4 H₂O(l) + 5 Fe³⁺(aq)

calculate how many grams of ammonia can be produced from 4.21 grams of

n2 + 3h2 = 2nh3

my answer:

25.0g n2 x (1 mol/ 28.0gn2) x (2 mol nh3/ 1 mol n2) x (17.0g nh3/ 1 molnh3) = 30.36g n2

25.0g h2 x (1 mol h2/ 2.00g h2) x (2 mol nh3/ 3 mol h2) x (17.0g nh3/ 1 mol nh3) = 141.67g h2

lr = n2

the maximum mass of nh3 is nh3 in grams