. Calculate the mass of O2 produced if 3.450 g potassium chlorate is completely decomposed by heating in presence of a catalyst
1 answer:
<span>2 KClO3(s) → 3 O2(g) + 2 KCl(s)
</span><span>Note: MnO2 (Manganese Dioxide) is not part of the reaction. A catalyst lowers the activation energy and increases both forward and reverse reactions at equal rates.
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molar mass of KClO3 = 122.5
Moles of KClO3 = 3.45 / 122.55 = 0.028
Moles of O2 produce =
= 0.042 moles
molar mass of O2 = 32
so, mass of O2 = 32 x 0.042 = 1.35 g
</span><span>Note: MnO2 (Manganese Dioxide) is not part of the reaction. A catalyst lowers the activation energy and increases both forward and reverse reactions at equal rates.
</span>
molar mass of KClO3 = 122.5
Moles of KClO3 = 3.45 / 122.55 = 0.028
Moles of O2 produce =
= 0.042 moles
molar mass of O2 = 32
so, mass of O2 = 32 x 0.042 = 1.35 g
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Answer:
3.81 g Pb
Explanation:
When a lead acid car battery is recharged, the following half-reactions take place:
Cathode: PbSO₄(s) + H⁺ (aq) + 2e⁻ → Pb(s) + HSO₄⁻(aq)
Anode: PbSO₄(s) + 2 H₂O(l) → PbO₂(s) + HSO₄⁻(aq) + 3H⁺ (aq) + 2e⁻
We can establish the following relations:
- 1 A = 1 c/s
- 1 mole of Pb(s) is deposited when 2 moles of e⁻ circulate.
- The molar mass of Pb is 207.2 g/mol
- 1 mol of e⁻ has a charge of 96468 c (Faraday's constant)
Suppose a current of 96.0A is fed into a car battery for 37.0 seconds. The mass of lead deposited is: