Sodium nitrite (NaNO2) of mass 4.14 g is treated with excess potassium iodide in acetic acid, 2NO2^- + 2I^- + 4H+ -> 2NO + I2 + 2H2O. Calculate the mass of iodine liberated in grams, assuming complete reaction. Give the answer to two decimal places. (Na 23, N 14, O 16, I 127)

JEE Main levelNumerical answerDifficulty 2 of 3qa178

Sodium nitrite ( $NaNO X_{2}$ ) of mass 4.14 g is treated with excess potassium iodide in acetic acid, $2 NO X_{2} X^{-} + 2 I X^{-} + 4 H X^{+} 2 NO + I X_{2} + 2 H X_{2} O$ . Calculate the mass of iodine liberated in grams, assuming complete reaction. Give the answer to two decimal places. (Na 23, N 14, O 16, I 127)

Numerical answer type, answer in g, tolerance 0.01.

Molar mass of

NaNO X_{2}

23 + 14 + 32 = 69 g mo l^{- 1}

, so moles of nitrite

= 4.14/69 = 0.060

. The equation gives 1 mol

I X_{2}

per 2 mol nitrite, so

n (I X_{2}) = 0.0300

mol. Mass

= 0.0300 \times 254 = 7.62