Question

The following unbalanced reaction takes place in acidic solution. S2−(aq)+NO3−(aq) → SO42−(aq)+NO(g)

How many milliliters of a 12.0 M HNO3 solution are needed to dissolve 25.3 g of CuS?

Answer 1

Given equation is: s? (aq) + Noz (aq) → sou (aa) 4 NO (9) +5 - 2. +62- -> sou S +2 > No Nos decrease in Increasein oxidation no coxidation oxidationing - Reduching oxidations 2- Stepis s sou No No Stepi: Balance ataus olher Step D! 7 Nos >> NO +2440 Stepat ataus. - 2- Stepos 2 2- 2- than oxygen and Hydrogen, s2- sol Stepas Balance cxygen Not4H1 NO+2420 3² + 4H₂O sou =1+4+3 Stepīt! Balance Hydrogen ataus, NO+ 4H+ + 36- s+ 44,0 5 so+ 8H7| - NO+2H2O -2 to - 278 - -2 Step to Balance charge s? 4H2O → S047847 tse 1 -01 0x3 = 35?712620 ~ 3504 + 24874256€ @x8 7 8 Noz + 32H++ zyé -> 8N0+1640 on Adding 35° 78107 + 8H4> 3504 +44207800 +6

Balanced equation in Acidic mediumu ixi 35" + 8 NOE + 8H+- 3504 + 44,04 8 NO 3 mole of cus 8 mole of HNO required, 25.3 g of cus no of moly of cus = Mass Molar mars 25. 38 95.5 g/mol = 0.265 8x mol = 0.265 mol Let the volume at Hno be: x rue no otmoles of HNO2 = 12.0 M) (1 6oo). = (12.01 mg) * (Yose) * = 123 ruol 3 mole of cus → smole of Hno required 0.265 molestus - ? ng of moles of HMO, regeeired = 0.265x8 This is equal to 12x (Ooo 0 265X8 12% - 0.265*8 12 % TOGO 3 x= 0.265*8*1000 - 58.88 3412 voluame Of Hno, solution = 58.88 ml