Answer:
B
Explanation:
Oxidation only occur at Anode and Reduction at Cathode
<span>A single-replacement reaction</span>
A) For balanced chemical equation: 2HgO(s) → 2Hg(l) + O₂(g).
1) Mole ratio 1: n(HgO) : n(Hg) = 2 : 2 (1 : 1).
2) Mole ratio 2: n(HgO) : n(O₂) = 2 : 1.
3) Mole ratio 3: n(Hg) : n(O₂) = 2 : 1.
B) Balanced chemical equation: 4NH₃(g) + 6NO(g) → 5N₂(g) + 6H₂O(l).
1) Mole ratio 1: n(NH₃) : n(NO) = 4 : 6 (2 : 3).
2) Mole ratio 2: n(NH₃) : n(N₂) = 4 : 5.
3) Mole ratio 3: n(NH₃) : n(H₂O) = 4 : 6 (2 : 3).
4) Mole ratio 4: n(NO) : n(N₂) = 6 : 5.
5) Mole ratio 5: n(NO) : n(H₂O) = 6 : 6 (1 :1).
6) Mole ratio 6: n(N₂) : n(H₂O) = 5 : 6.
What is this?????????? looks cool
Convert all the grams to moles first.
K - 3.65 g / 39.1 g/mol = 0.0934 mol
Cl - 3.33 g / 35.45 g/mol = 0.0939 mol
O - 3.02 g / 16 g/mol = 0.189 mol
Divide them all by the smallest number of moles (0.0934 moles of K) to get whole numbers.
K - 0.0934/0.0934 = 1
Cl - 0.0939/0.0934 ≈ 1
O - 0.189/0.0934 ≈ 2
KClO2
