In the brown ring test, $3 Fe X^{2 +} + NO X_{3} X^{-} + 4 H X^{+} 3 Fe X^{3 +} + NO + 2 H X_{2} O$ . A mixture of $2.28 g$ of $FeSO X_{4}$ and $1.01 g$ of $KNO X_{3}$ is treated in excess dilute acid with the reaction proceeding as written. Taking Fe 56, S 32, O 16, K 39, N 14 and the molar volume at STP as $22.4 L mo l^{- 1}$ , find the volume of NO formed at STP in mL, to the nearest integer.

Numerical answer type, answer in mL, tolerance 0.5.

Answer: 112 mL

M (FeSO X_{4}) = 152

, so

Fe X^{2 +} = 2.28/152 = 0.0150 mol

;

M (KNO X_{3}) = 101

, so nitrate

= 0.0100 mol

. Iron needs 3 mol per mol nitrate, so

Fe X^{2 +}

is limiting: NO

= 0.0150/3 = 0.00500 mol

(nitrate would allow

0.0100

). Volume

= 0.00500 \times 22400 = 112 mL