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: B. 0.05 mol and 0.15 mol
- A. 0.1 mol and 0.3 mol
- B. 0.05 mol and 0.15 mol
- C. 0.02 mol and 0.06 mol
- D. 0.1 mol and 0.15 mol
Explanation
To find the correct amounts of KIO3 consumed and ICl formed, first, the balanced chemical equation must be established. Based on stoichiometry, the moles of KI can be determined from its mass:Molar mass of KI = 166 g/mol, thus:Moles of KI = 16.6 g / 166 g/mol = 0.1 mol.Assuming the balanced reaction shows that 1 mole of KI reacts with 1 mole of KIO3 to produce 3 moles of ICl, the limiting reactant is KI, and we would need only 0.05 mol of KIO3 (0.1 mol KI) to react completely with it. Therefore, the amount of ICl produced would be 0.15 mol (0.1 mol KI produces 0.3 mol ICl). Hence, the correct answer is Option B: 0.05 mol of KIO3 consumed and 0.15 mol of ICl formed. The other options either overestimate or underestimate the amounts based on the stoichiometric relationships established in the balanced equation.
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Mole calculations connect mass, number of particles and Avogadro's number, while balanced equations provide the mole ratios used in stoichiometry. Questions cover limiting and excess reactants, theoretical yield and percentage yield, including identifying which reactant is consumed first and comparing the actual product with the maximum possible product.
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