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Limiting Reagent or Reactant

2PbS+3O2→ 2Pb...

Question

$2 P b S + 3 O_{2} \to 2 P b O + 2 S O_{2}$ ;
$3 S O_{2} + 2 H N O_{3} + 2 H_{2} O \to 3 H_{2} S O_{4} + 2 N O$

According to the above sequence of reactions, how much $H_{2} S O_{4}$ will $1200$ g of $P b S$ produce?

[Molar mass of $P b =$ $208$ ]

490

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520

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560

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580

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Solution

The correct option is B $490$ g
$2 P b S + 3 O_{2} \to 2 P b O + 2 S O_{2}$ ;
$3 S O_{2} + 2 H N O_{3} + 2 H_{2} O \to 3 H_{2} S O_{4} + 2 N O$

From the reaction, it can be seen that $1$ mole of $P b S$ will produce $1$ mole of acid.

So, $\frac{1200}{240} = 5$ moles will produce $98 \times 5 = 490$ g $H_{2} S O_{4}$ .

Suggest Corrections

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{2PbS + 3O}_{2} \to 2 {PbO + 2SO}_{2}

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H_{2} {SO}_{4}

P b S

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 %)

550 kg

P b S

P b

P b = 207 g/mol

2 P b S + 3 O_{2} \to 2 P b O + 2 S O_{2}

P b S + 2 O_{2} \to P b S O_{4}

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438 k J

22.3 g P b O (s)

P b = 207, O = 16.0)

2 P b (s) + O_{2} (g) \to 2 P b O (s)

N_{2}

O_{2}

N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)

N_{2}

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