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NaNO(2) + NaI + H(2) SO(4) = NO + I(2) +...

`NaNO_(2) + NaI + H_(2) SO_(4) = NO + I_(2) + Na_(2) SO_(4) + H_(2)O`

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To balance the redox reaction given by: \[ \text{NaNO}_2 + \text{NaI} + \text{H}_2\text{SO}_4 \rightarrow \text{NO} + \text{I}_2 + \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} \] we will follow these steps: ### Step 1: Identify Oxidation States - In NaNO₂, the oxidation state of nitrogen (N) is +3. ...
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HNO_(2) + KMnO_(4) + H_(2) SO_(4) = HNO_(3) + KMnO_(4) + K_(2) SO_(4) + H_(2)O

Na_(2)O_(2) + H_(2)SO_(4) to

Knowledge Check

  • The equiv. wt. of hypo in the reaction Na_(2) S_(2) O_(3) + Cl_(2) + H_(2) O rarr Na_(2) SO_(4) + H_(2) SO_(4) + HCl is -

    A
    `("Mol. wt")/(2)`
    B
    `("Mol. wt")/(4)`
    C
    `("Mol. wt")/(1)`
    D
    `("Mol. wt")/(8)`

  • The equiv. wt. of hypo in the reaction Na_(2) S_(2) O_(3) + Cl_(2) + H_(2) O rarr Na_(2) SO_(4) + H_(2) SO_(4) + HCl is -

    A
    `( "Mol. Wt. ")/( 2) `
    B
    `( "Mol. Wt. ")/( 4) `
    C
    `( "Mol. Wt. ")/( 1) `
    D
    `( "Mol. Wt. ")/( 8) `

  • In the reactions HCI + NaOH rightarrow NaCI + H_(2)O + x cal. H_(2)SO_(4) + 2NaOh rightarrow Na_(2)SO_(4) + 2 H_(2)O + y cal.

    A
    ` x = y `
    B
    `x = 2y`
    C
    `x = y/2`
    D
    `x = sqrt y`

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    Identify oxidising and reducing agent in equation Na NO _(2) + KI + H _(2) SO _(4) to Na _(2) SO _(4) + K_(2) SO _(4) + NO + H _(2) O + I _(2)

    Balance : Na_(2) S_(2) O_(3) + KMnO_(4) + H_(2) O to K_(2) SO_(4) + Na_(2) SO_(4) + KOH + MnO_(2)

    Balance the following equations : (i) KMnO_(4) + H_(2)SO_(4) + H_(2)O_(2) to K_(2)SO_(4) + MnSO_(4) + H_(2)O + O_(2) (ii) KMnO_(4) + KCl + H_(2)SO_(4) to MnSO_(4) + K_(2)SO_(4) + H_(2)O + Cl_(2) (iii) MnO_(2) + H_(2)O_(2) to MnO_(4)^(-) + MnO_(4)^(-) + H_(2)O (Basic medium)

    Calculate the equivalent weight of each oxidant and reducant in: (a) FeSO_(4) + KCl_(3) rarr KCl + Fe_(2) (SO_(4))_(3) (b) Na_(2) SO_(3) + Na_(2)CrO_(4) rarr Na_(2) CrO_(4) rarr Na_(2) SO_(4) + Cr(OH)_(3) (c) Fe_(3)O_(4) + KMnO_(4) rarr Fe_(2) O_(3) + MnO_(2) (d) KI + K_(2) Cr_(2) O_(7) rarr Cr^(3+) + 3I_(2) (e) Mn^(4+) rarr Mn^(2+) (f) NO_(3)^(-) rarr N_(2) (g) N_(2) rarr NH_(3) (h) Na_(2) S_(2) O_(3) + I_(2) rarr Na_(2) S_(4) O_(6) + 2NaI (i) FeC_(2)O_(4) rarr Fe^(3+) + CO_(2)

    Balance : As_(2) S_(3) + NaCl O_(3) to Na_(2) As O_(4) + Na_(2) SO_(4) + NaClO + H_(2) O

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