IDNLearn.com: Where your questions are met with thoughtful and precise answers. Join our platform to receive prompt and accurate responses from experienced professionals in various fields.

Determine the theoretical mass, in grams, of lithium nitride if 5.0 grams of lithium reacts in an excess of nitrogen gas.

Sagot :

Certainly! Let's walk through the solution step-by-step.

### Step 1: Determine the molar masses
- Molar mass of lithium (Li): 6.94 grams per mole
- Molar mass of lithium nitride (Li₃N): 34.83 grams per mole

### Step 2: Calculate the moles of lithium (Li)
Given the mass of lithium is 5.0 grams:
[tex]\[ \text{Moles of Li} = \frac{\text{Mass of Li}}{\text{Molar mass of Li}} \][/tex]
[tex]\[ \text{Moles of Li} = \frac{5.0 \text{ grams}}{6.94 \text{ grams/mole}} \][/tex]
[tex]\[ \text{Moles of Li} \approx 0.7205 \text{ moles} \][/tex]

### Step 3: Utilize the stoichiometry of the reaction
The reaction between lithium and nitrogen gas to form lithium nitride is:
[tex]\[ 6 \text{ Li} + \text{N}_2 \rightarrow 2 \text{ Li}_3\text{N} \][/tex]
From this stoichiometry, 6 moles of lithium produce 2 moles of lithium nitride, which simplifies to:
[tex]\[ 3 \text{ moles of Li} \rightarrow 1 \text{ mole of Li}_3\text{N} \][/tex]

### Step 4: Calculate the moles of lithium nitride (Li₃N)
Using the ratio from the reaction, the moles of Li₃N formed can be calculated as follows:
[tex]\[ \text{Moles of Li₃N} = \frac{\text{Moles of Li}}{3} \][/tex]
[tex]\[ \text{Moles of Li₃N} = \frac{0.7205 \text{ moles}}{3} \][/tex]
[tex]\[ \text{Moles of Li₃N} \approx 0.2402 \text{ moles} \][/tex]

### Step 5: Calculate the mass of lithium nitride (Li₃N)
Finally, we need to find the mass of lithium nitride by using its molar mass:
[tex]\[ \text{Mass of Li₃N} = \text{Moles of Li₃N} \times \text{Molar mass of Li₃N} \][/tex]
[tex]\[ \text{Mass of Li₃N} = 0.2402 \text{ moles} \times 34.83 \text{ grams/mole} \][/tex]
[tex]\[ \text{Mass of Li₃N} \approx 8.36 \text{ grams} \][/tex]

Therefore, the theoretical mass of lithium nitride formed when 5.0 grams of lithium reacts in excess nitrogen gas is approximately 8.36 grams.
We are delighted to have you as part of our community. Keep asking, answering, and sharing your insights. Together, we can create a valuable knowledge resource. Discover insightful answers at IDNLearn.com. We appreciate your visit and look forward to assisting you again.