When 16.6 g of KI is treated with an excess of KIO3 in the presence of 6NHCl, ICl is produced. The amount of KIO3 consumed and the ICl formed are:
Correct answer: C. 0.05 mol and 0.15 mol
- A. 0.1 mol and 0.3 mol
- B. 0.05 mol and 0.3 mol
- C. 0.05 mol and 0.15 mol
- D. 0.1 mol and 0.15 mol
Explanation
First, form a balanced equation of the reaction.KIO3 + 2 KI + 6 HCl → 3 ICl + 3 KCl + 3 H2OCalculate the moles of KI used Mass of KI = 16.6g and Mr of KI = 39 + 127 = 166 Moles = mass/mr = 16.6/166 = 0.1 molesMolar ratio of KI to KIO3 KI : KIO32 : 1 As seen in the equation. So, according to this every 2 moles of KI need 1 moles of KIO3. So, 0.1 moles of KI will need 0.05 moles of KIO3. (As per the ratio)(1/2) x 0.1 = 0.05 molesSimilarly Molar ratio of KI and ICl is KI : ICL2 : 3 As seen in the equation. So, according to this every 2 moles of KI produce 3 moles of ICl. So, o.1 moles of KI will produce o,15 moles of ICl. (As per the ratio)(3/2) x 0.1 = 0.15 moles.Hence, moles of KIO3 = 0.05 moles ICl = 0.15 moles
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