The number of ions present in 2.0 L of a solution of 0.8 M K4 [Fe(CN)6] is:
Correct answer: B. 4.8 x 10^24
- A. 4.8 x 10^22
- B. 4.8 x 10^24
- C. 9.6 x 10^24
- D. 9.6 x 10^22
Explanation
Step 1: Dissociation of the compoundK₄[Fe(CN)₆] is an ionic compound and dissociates in water as:K4 [Fe(CN)6 ]→4K++[Fe(CN)6 ]4− So, 1 mole of K₄[Fe(CN)₆] produces 5 ions in total:Step 2: Calculate moles of K₄[Fe(CN)₆]Moles=Molarity×Volume (in L)=0.8mol/L×2.0L=1.6mol Step 3: Total number of ionsSince each mole gives 5 ions:Total moles of ions=1.6mol×5=8.0mol of ions Step 4: Convert moles to number of ionsUse Avogadro's number:1mol=6.022×1023particles Number of ions=8.0×6.022×1023=4.818×1024ions
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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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