Problem: Consider the following reaction:4 K(s) + O2 (g) → 2 K2O(s)The molar mass of K is 39.09 g/mol and that of O2 is 32.00 g/mol. Without doing any calculations, pick the conditions under which potassium is the limiting reactant. Explain your reasoning.a) 16 g K, 2.5 g O2 b) 1.5 g K, 0.38 g O2c) 165 g K, 28 g O2d) 170 g K, 31 g O2

FREE Expert Solution

Balanced equation: 4 K(s) + O2 (g) → 2 K2O(s)

  • Mole to mole comparison: 
    • 4 mol K reacts with 1 mole O
    • Limiting reactant → produces least amount of product → gets used up when the reaction is complete. 
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Problem Details

Consider the following reaction:

4 K(s) + O2 (g) → 2 K2O(s)

The molar mass of K is 39.09 g/mol and that of O2 is 32.00 g/mol. Without doing any calculations, pick the conditions under which potassium is the limiting reactant. Explain your reasoning.


a) 16 g K, 2.5 g O2 

b) 1.5 g K, 0.38 g O2

c) 165 g K, 28 g O2

d) 170 g K, 31 g O2

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Our tutors have indicated that to solve this problem you will need to apply the Limiting Reagent concept. If you need more Limiting Reagent practice, you can also practice Limiting Reagent practice problems.