हिंदी

1.56 g of sodium peroxide reacts with water according to the following equation: 2NaA2OA2+2HA2O⟶4NaOH+OA2 Calculate the volume of oxygen liberated at STP.

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प्रश्न

1.56 g of sodium peroxide reacts with water according to the following equation:

\[\ce{2Na2O2 + 2H2O -> 4NaOH + O2}\]

Calculate the volume of oxygen liberated at STP.

संख्यात्मक
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उत्तर

156 g of Na2O2 liberates oxygen at S.T.P. = 22.4 L

∴ 1.56 g of Na2O2 liberates oxygen

= `22.4/156 xx 1.56`

= 0.224 L

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अध्याय 5: Mole concept and Stoichiometry - EXERCISE-5D [पृष्ठ ९४]

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एस पी सिंग Concise Chemistry [English] Class 10 ICSE
अध्याय 5 Mole concept and Stoichiometry
EXERCISE-5D | Q 8. (b) | पृष्ठ ९४
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