Question

Consider the following reaction:

4K(s)+O2(g)→2K2O(s)4K(s)+O2(g)→2K2O(s)
The molar mass of KK is 39.10 gmol−1gmol−1 and that of O2O2 is 32.00 gmol−1gmol−1.
Consider the following reaction:
4K(s)+O₂(g)→2K₂O(s)4K(s)+O₂(g)→2K₂O(s)
The molar mass of KK is 39.10 gmol−1gmol−1 and that of O2O2 is 32.00 gmol−1gmol−1.Without doing any calculations, pick the conditions under which potassium is the limiting reactant
A. 170 g K, 31g O₂
B. 165 kg K, 28 kg O₂
C. 16 g K, 2.5 g O₂
D. 1.5g K, 0.38 g O₂

Answer 1

The limiting reactant when given 1.5g of potassium (K) and 0.38g of oxygen (O₂) is potassium (K), because it is present in a smaller mole ratio than required by the balanced reaction.

Step-by-step explanation:

To determine the limiting reactant in a chemical reaction, we must compare the mole ratio of the reactants to the coefficients in the balanced chemical equation. The reaction is 4K(s) + O₂(g) → 2K₂O(s), which means that 4 moles of K reacts with 1 mole of O₂.

To find the limiting reactant without doing calculations, we need to find the option where potassium (K) is present in less than the stoichiometric ratio compared to oxygen (O₂). The molar mass of K is 39.10 g/mol, and O₂ is 32.00 g/mol. We can observe that for every 4 moles of K (which would be 4 x 39.10 g = 156.4 g), we need 1 mole of O₂ (32 g).

By comparing the given mass of each reactant to these amounts, we see that in option D, 1.5 g K is far less than the ratio needs (156.4 g), and 0.38 g O₂ is also less, but proportionally more compared to K needed for the complete reaction. Therefore, in option D, K is the limiting reactant.