Lithium oxide can be produced by the reaction shown below. If a manufacturing company wants to make 25.0 g of Li2O, what mass of Li is necessary? [Molar masses: Li, 6.94 g/mol; O, 16.00 g/mol] 4Li(s) + O2(g) → 2Li2O(s) A) 0.0861 g Li B) 2.903 g Li C) 11.6 g Li D) 12.5 g Li E) 23.2 g Li
The Correct Answer and Explanation is:
Correct Answer: C) 11.6 g Li
Step by Step Explanation:
Step 1: Write the balanced equation4Li(s)+O2(g)→2Li2O(s)4Li (s) + O_2 (g) \rightarrow 2Li_2O (s)4Li(s)+O2(g)→2Li2O(s)
Step 2: Find the molar masses
- Lithium (Li) = 6.94 g/mol
- Oxygen (O) = 16.00 g/mol
- Lithium oxide (Li₂O) = (2×6.94)+16.00=13.88+16.00=29.88 g/mol(2 \times 6.94) + 16.00 = 13.88 + 16.00 = 29.88 \, g/mol(2×6.94)+16.00=13.88+16.00=29.88g/mol
Step 3: Calculate the moles of Li₂O required
We are asked to produce 25.0 g of Li₂O. Using molar mass:Moles of Li2O=25.0 g29.88 g/mol=0.8365 mol\text{Moles of } Li_2O = \frac{25.0 \, g}{29.88 \, g/mol} = 0.8365 \, molMoles of Li2O=29.88g/mol25.0g=0.8365mol
Step 4: Use the mole ratio from the equation
From the balanced reaction:
- 4 moles of Li produce 2 moles of Li₂O
- Therefore, to produce 0.8365 moles of Li₂O:
Moles of Li needed=0.8365 mol×4 mol Li2 mol Li2O=1.673 mol Li\text{Moles of Li needed} = 0.8365 \, mol \times \frac{4 \, mol \, Li}{2 \, mol \, Li_2O} = 1.673 \, mol \, LiMoles of Li needed=0.8365mol×2molLi2O4molLi=1.673molLi
Step 5: Calculate mass of Lithium requiredMass of Li=1.673 mol×6.94 g/mol=11.61 g\text{Mass of Li} = 1.673 \, mol \times 6.94 \, g/mol = 11.61 \, gMass of Li=1.673mol×6.94g/mol=11.61g
Final Mass Rounded to Three Significant Figures11.6 g11.6 \, g11.6g
Conclusion:
The company needs 11.6 grams of Lithium to produce 25.0 grams of Lithium Oxide.
Thus, the correct choice is C) 11.6 g Li.
