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Limiting Reagent or Reactant

If 0.740 g ...

Question

If $0.740$ g of $O_{3}$ reacts with $0.670$ g of $N O$ , how many grams of $N O_{2}$ will be produced?

0.71

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0.74

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0.68

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0.81

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Solution

The correct option is A $0.71$ g
$0.74$ g $O_{3} = \frac{0.74}{48} = 0.0154$ mol $O_{3}$
$0.67$ g $N O = \frac{0.67}{30} = 0.0223$ mol $N O$
$O_{3}$ is the limiting reagent and NO is in excess $= 0.0223 - 0.0154 = 0.007$ mol
Thus, $O_{3}$ taken $= N O_{2}$ formed $= 0.0154$ mol $N O_{2}$
$= 0.0154 \times 46$ g $N O_{2}$
$= 0.71$ g

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Similar questions

O_{3} + N O ⟶ O_{2} + N O_{2}

0.740

O_{3}

0.670

N O

N O_{2}

60

N O

O_{2}

N O_{2}

N O_{2}

N O

= 30 g / m o l

N O_{2}

= 46 g / m o l

{N a}_{2} {S O}_{4}

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H_{2} {S O}_{4}

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N a O H

10 g

K H C O_{3}

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K_{2} C O_{3}

1.07 g

H_{2} O

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