A cyanide leach solution contains 4.925 g of gold as Na[Au(CN)2]. Zinc dust is added by the reaction 2Na[Au(CN)2] + Zn -> Na2[Zn(CN)4] + 2Au and 91.0% of the gold is precipitated. Taking Au = 197 and Zn = 65, find the mass of zinc consumed, in mg (nearest integer).

JEE Main levelNumerical answerDifficulty 2 of 3mt028

A cyanide leach solution contains $4.925 g$ of gold as $Na [Au (CN) X_{2}]$ . Zinc dust is added by the reaction $2 Na [Au (CN) X_{2}] + Zn Na X_{2} [Zn (CN) X_{4}] + 2 Au$ and $91.0%$ of the gold is precipitated. Taking $Au = 197$ and $Zn = 65$ , find the mass of zinc consumed, in mg (nearest integer).

Numerical answer type, answer in mg, tolerance 0.5.

Moles of Au

= 4.925/197 = 0.0250

. Precipitated Au

= 0.91 \times 0.0250 = 0.02275

mol. The equation needs 1 mol Zn per 2 mol Au, so Zn

= 0.011375

mol. Mass

= 0.011375 \times 65 = 739.4 mg

i.e. 739 mg to the nearest integer.