Question

Use the chemical equation to answer the question.
4K(s) + O2 (g) → 2K O(s)
The molar mass of potassium (K) is 39 g/mol. The molar mass of oxygen (O) is 16 g/mol. How many grams of potassium are needed in
the reaction to produce 23.5 grams of potassium oxide (K O)?
(1 point)
2 2
2
9.75 g
0.50 g
23.5 g
19.5 G

Answer 1

To solve this problem, we first need to determine the molar mass of potassium oxide (K₂O).

1. Calculate the molar mass of potassium oxide (K₂O):
- The molar mass of potassium (K) = 39 g/mol
- The molar mass of oxygen (O) = 16 g/mol
- Molar mass of K₂O = (2 × 39 g/mol) + (1 × 16 g/mol) = 78 g/mol + 16 g/mol = 94 g/mol

2. Now, we need to determine how many moles of K₂O are in 23.5 grams:
\[
\text{Moles of K₂O} = \frac{\text{Mass of K₂O}}{\text{Molar mass of K₂O}} = \frac{23.5 \text{ g}}{94 \text{ g/mol}} \approx 0.250 \text{ moles}
\]

3. Next, we utilize the stoichiometry of the balanced chemical equation:
\[
4 \text{ moles of K} + 1 \text{ mole of O}_2 \rightarrow 2 \text{ moles of K}_2O
\]
From the equation, 2 moles of K₂O require 4 moles of K. Thus:
\[
\text{Moles of K needed} = 4 \text{ moles of K} \times \left(\frac{0.250 \text{ moles of K}_2O}{2 \text{ moles of K}_2O}\right) = 0.500 \text{ moles of K}
\]

4. Now calculate the mass of potassium (K) required:
\[
\text{Mass of K} = \text{Moles of K} \times \text{Molar mass of K} = 0.500 \text{ moles} \times 39 \text{ g/mol} = 19.5 \text{ g}
\]

Therefore, **19.5 g** of potassium are needed in the reaction to produce 23.5 grams of potassium oxide (K₂O).

So the answer is **19.5 g**.