Answer:
[tex]\boxed{\text{6.937 u; X = Li}}[/tex]
Explanation:
1. Write the unbalanced equation
X + N₂ ⟶ X₃N
2. Balance the equation and gather all the data.
MM: 28.01
6X + N₂ ⟶ 2X₃N
m/g 1.486 1.000
3. Calculate the moles of N₂
[tex]\text{Moles of N}_{2} = \text{1.000 g N}_{2} \times \dfrac{\text{1 mol N}_{2}}{\text{28.01 g N}_{2}} = \text{0.035 70 mol N}_{2}[/tex]
4. Calculate the moles of X
The molar ratio is 6 mol X: 1 mol N₂
[tex]\text{Moles of X} = \text{0.035 70 mol N}_{2} \times \dfrac{\text{6 mol X}}{\text{1 mol N}_{2}} = \text{0.2142 mol X}[/tex]
5. Calculate the molar mass of X
[tex]\text{Molar mass} = \dfrac{\text{mass}}{\text{moles}} = \dfrac{\text{1.486 g}}{\text{0.2142 mol}} = \text{6.937 g/mol}\\\text{The molar mass of X is 6.937 g/mol, so the atomic mass $\boxed{\textbf{6.937 u}}$}[/tex]
6. Identify X.
[tex]\text{X has an atomic mass of 6.937 u, so $\boxed{\textbf{X = Li}}$ (at. mass 6.94 u)}[/tex]
