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Limiting Reagent or Reactant

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Question

Calculate the weight of nitric acid present in $40 m l$ of solution which completely neutralizes $20 m l$ of $0.01 M$ barium hydroxide assuming complete ionization.

0.40 g

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0.0004 g

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0.0252 g

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0.063 g

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Solution

The correct option is C $0.0252 g$
equation is $2 H N O_{3} + B a (O H)_{2} ⇌ H_{2} O + B a (N O_{3})_{2}$
$M o l e o f B a (O H)_{2} = 0.01 M \times 0.02 l = 0.0002 m o l e$
1 mole of $B a (O H)_{2}$ need 2 moles of $H N O_{3}$ .
therefore $0.0002 m o l e$ will be neutralized by $0.0002 \times 2 = 0.0004 m o l e s H N O_{3}$
Weight of 0.004 mole $H N O_{3} = m o l e \times m o l e c u l a r w e i g h t$ = $0.0004 \times 63 = 0.0252 g$

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