Lead is formed from its ore by the following reactions- 2 PbS s -3 O - 2 g - rightarrow 2 PbO s -2 SO - 2 g -text Yield -40 - PbO s - C s rightarrow Pb s - CO g -text Yield -80 -If 550 kg of PbS was initially taken- find the amount of P b in kg formed- Molar mass of Pb -207 g - mol A- 147-75 kgB- 288-63 kgC- 152-42 kgD- 108-23 kg

The correct option is C 152.42 kg2PbS(s) + 3O_2 → 2PbO + 2SO_2 (i) PbO + C → Pb + CO (ii) Mol of PbS = nbsp;550239× 1000 = 2.301×1000 = 2301 moles As pe ...

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Standard XII

Chemistry

Stoichiometric Calculations

Lead is forme...

Question

Lead is formed from its ore by the following reactions: $2 P b S (s) + 3 O_{2} (g) \to 2 P b O (s) + 2 S O_{2} (g) (Yield = 40 %)$
$P b O (s) + C (s) \to P b (s) + C O (g) (Yield = 80 %)$
If $550 kg$ of $P b S$ was initially taken, find the amount of $P b$ in kg formed.
(Molar mass of $P b = 207 g/mol$ )

147.75 kg

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288.63 kg

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152.42 kg

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108.23 kg

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Solution

The correct option is C $152.42 kg$
$2 P b S (s) + 3 O_{2} \to 2 P b O + 2 S O_{2} (i)$
$P b O + C \to P b + C O (ii)$
Mol of $P b S = \frac{550}{239} \times 1000 = 2.301 \times 1000 = 2301 moles$
As per the stoichiometry of the reaction,
2 mol of $P b S$ produces 2 moles of $P b O$
2301 mol of $P b S$ will form 2301 mol of $P b O$
Yield of the reaction is given as 40 %
Actual amount of $P b O$ formed $= 2301 \times 0.40 = 920.4$ moles
In reaction (ii),
1 mol of $P b O$ produces 1 mol of $P b$
920.4 mol will form 920.4 mol of $P b$
Yield of the reaction is given as 80 %
Actual amount of Pb formed $= 920.4 \times 0.8 = 736.32$ moles
Amount of $P b$ formed $= 736.32 \times 207 = 152.42 kg$

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Lead is formed from its ore by the following reactions: 2PbS(s)+3O2(g)→2PbO(s)+2SO2(g) (Yield = 40 %) PbO(s)+C(s)→Pb(s)+CO(g) (Yield = 80 %) If 550 kg of PbS was initially taken, find the amount of Pb in kg formed. (Molar mass of Pb=207 g/mol)