Answer :
Answer:
[tex]1.86\times10^{23}\text{ atoms Cu}[/tex]
Explanation:
First, we can solve for how many moles of Nitrogen gas will react using the volume of an ideal gas at STP (22.41 L):
[tex]n = \dfrac{41.64\text{ L}}{1}\times\dfrac{1\text{ mole}}{22.41\text{ L}} \approx 1.858\text{ mole N}_2[/tex]
Next, we can multiply this by the stoichiometric ratio given by the reaction equation (which we have to balance):
[tex]\rm 6\,Cu+N_2\to 2\,Cu_3N[/tex]
↓↓↓
[tex]\rm \dfrac{1.858\text{ mole N}_2}{1}\times \dfrac{6\text{ mole Cu}}{1\text{ mole N}_2} \times\dfrac{6.022\times10^{23}\text{ atoms Cu}}{1\text{ mole Cu}} \approx 1.8649\times10^{23}\text{ atoms Cu}[/tex]
Finally, we can round to 2 decimal places:
[tex]\boxed{1.86\times10^{23}\text{ atoms Cu}}[/tex]
