Question
2 G of potassium (K) reacts with 5 G of oxygen (o) according to the law of conservation of mass how many grams of potassium oxide (K20) will be produced
A. 5 grams
B. 10 grams
C. 7 grams
D. 2 grams
A. 5 grams
B. 10 grams
C. 7 grams
D. 2 grams
Answers
GPT-4o mini
To find out how many grams of potassium oxide (K₂O) will be produced from the reaction of potassium (K) with oxygen (O), we can start by using the law of conservation of mass. This law states that mass is neither created nor destroyed in a chemical reaction; thus, the total mass of the reactants will equal the total mass of the products.
1. **Calculate the total mass of the reactants:**
- Mass of potassium (K) = 2 g
- Mass of oxygen (O) = 5 g
- Total mass of reactants = \(2 \, \text{g} + 5 \, \text{g} = 7 \, \text{g}\)
2. **Determine the products:**
The chemical reaction for the formation of potassium oxide (K₂O) can be represented as:
\[
4 \, \text{K} + \text{O}_2 \rightarrow 2 \, \text{K}_2\text{O}
\]
To find out how much K₂O can be produced:
- From the balanced equation:
- 4 moles of potassium combine with 1 mole of oxygen to produce 2 moles of potassium oxide.
- The molar mass of K = 39 g/mol, so 2 g of K is:
- \(\frac{2 \, \text{g}}{39 \, \text{g/mol}} \approx 0.0513 \, \text{moles of K}\)
- The molar mass of O₂ = 32 g/mol, so 5 g of O₂ is:
- \(\frac{5 \, \text{g}}{32 \, \text{g/mol}} \approx 0.15625 \, \text{moles of O}_2\)
3. **Inspect the limiting reactant:**
- From the reaction stoichiometry, 4 moles of K react with 1 mole of O₂. Therefore:
- 0.0513 moles of K would require \(0.0513 \times \frac{1}{4} \approx 0.0128\) moles of O₂ to completely react.
- Since 0.15625 moles of O₂ are available, K is the limiting reactant.
4. **Calculate the amount of K₂O produced:**
- From the balanced equation, 4 moles of K produce 2 moles of K₂O, hence:
- \(0.0513 \, \text{moles of K} \times \frac{2}{4} = 0.02565 \, \text{moles of K}_2\text{O}\)
5. **Convert moles of K₂O to grams:**
- The molar mass of K₂O = (39*2) + 16 = 94 g/mol, hence:
- Mass of K₂O = \(0.02565 \, \text{moles} \times 94 \, \text{g/mol} \approx 2.41 \, \text{g}\)
Since the total mass of the reactants was 7 g, we understand the entire mass is conserved.
Thus, the closest answer according to the initial provided options is:
C. 7 grams
But given the calculated value, the expected outcome seems like a discrepancy in a complete reaction where some theoretical assumptions are applied, but from the options:
C. 7 grams.
1. **Calculate the total mass of the reactants:**
- Mass of potassium (K) = 2 g
- Mass of oxygen (O) = 5 g
- Total mass of reactants = \(2 \, \text{g} + 5 \, \text{g} = 7 \, \text{g}\)
2. **Determine the products:**
The chemical reaction for the formation of potassium oxide (K₂O) can be represented as:
\[
4 \, \text{K} + \text{O}_2 \rightarrow 2 \, \text{K}_2\text{O}
\]
To find out how much K₂O can be produced:
- From the balanced equation:
- 4 moles of potassium combine with 1 mole of oxygen to produce 2 moles of potassium oxide.
- The molar mass of K = 39 g/mol, so 2 g of K is:
- \(\frac{2 \, \text{g}}{39 \, \text{g/mol}} \approx 0.0513 \, \text{moles of K}\)
- The molar mass of O₂ = 32 g/mol, so 5 g of O₂ is:
- \(\frac{5 \, \text{g}}{32 \, \text{g/mol}} \approx 0.15625 \, \text{moles of O}_2\)
3. **Inspect the limiting reactant:**
- From the reaction stoichiometry, 4 moles of K react with 1 mole of O₂. Therefore:
- 0.0513 moles of K would require \(0.0513 \times \frac{1}{4} \approx 0.0128\) moles of O₂ to completely react.
- Since 0.15625 moles of O₂ are available, K is the limiting reactant.
4. **Calculate the amount of K₂O produced:**
- From the balanced equation, 4 moles of K produce 2 moles of K₂O, hence:
- \(0.0513 \, \text{moles of K} \times \frac{2}{4} = 0.02565 \, \text{moles of K}_2\text{O}\)
5. **Convert moles of K₂O to grams:**
- The molar mass of K₂O = (39*2) + 16 = 94 g/mol, hence:
- Mass of K₂O = \(0.02565 \, \text{moles} \times 94 \, \text{g/mol} \approx 2.41 \, \text{g}\)
Since the total mass of the reactants was 7 g, we understand the entire mass is conserved.
Thus, the closest answer according to the initial provided options is:
C. 7 grams
But given the calculated value, the expected outcome seems like a discrepancy in a complete reaction where some theoretical assumptions are applied, but from the options:
C. 7 grams.