A scientist wants to determine how many moles of sodium nitride $(Na_3N)$ could be made from 7.83 mol of nitrogen $(N_2)$.
$6 Na+N_2 \rightarrow 2 Na_3 N$
How many moles of sodium nitride could be made?
Answer (1)
Determine the mole ratio between N 2 and N a 3 N from the balanced equation: 6 N a + N 2 → 2 N a 3 N .
Use the mole ratio to convert moles of N 2 to moles of N a 3 N .
Multiply the moles of N 2 (7.83 mol) by the mole ratio (2 mol N a 3 N / 1 mol N 2 ).
Calculate the moles of N a 3 N : 7.83 × 2 = 15.66 mol. The final answer is 15.66
Explanation
Problem Setup and Given Information We're given a chemical reaction where nitrogen gas ( N 2 ) reacts with sodium to produce sodium nitride ( N a 3 N ). The balanced equation is:
6 N a + N 2 → 2 N a 3 N
We are told that we have 7.83 moles of N 2 and we want to find out how many moles of N a 3 N can be produced.
Determining the Mole Ratio From the balanced equation, we can see that 1 mole of N 2 produces 2 moles of N a 3 N . This gives us a mole ratio of 2 moles N a 3 N / 1 mole N 2 . We can use this ratio to convert moles of N 2 to moles of N a 3 N .
Calculating Moles of Sodium Nitride To find the moles of N a 3 N produced, we multiply the moles of N 2 by the mole ratio:
Moles of N a 3 N = (Moles of N 2 ) × (Mole ratio)
Moles of N a 3 N = 7.83 mol N 2 × 1 mol N 2 2 mol N a 3 N
Final Calculation Now, let's do the calculation:
Moles of N a 3 N = 7.83 × 2 = 15.66 moles
Final Answer Therefore, 7.83 moles of N 2 can produce 15.66 moles of N a 3 N .
Examples
In the manufacturing of semiconductors, precise chemical reactions are crucial. For instance, if you're producing silicon nitride layers using nitrogen gas and silane, knowing the exact stoichiometric relationships helps control the deposition process. If a process requires 7.83 moles of nitrogen gas, understanding the reaction's stoichiometry allows engineers to calculate precisely how much silicon nitride can be produced, ensuring optimal material usage and product quality. This precise control minimizes waste and maximizes the efficiency of the manufacturing process, directly impacting cost and yield.
