Answer:
26.95 %
Explanation:
Air contains the highest percentage of oxygen and nitrogen gases. Magnesium then combines with both of the gases:
[tex]2 Mg (s) + O_2 (g)\rightarrow 2 MgO (s)[/tex]
[tex]3 Mg (s) + N_2 (g)\rightarrow Mg_3N_2 (s)[/tex]
Firstly, find the total number of moles of magnesium metal:
[tex]n_{Mg} = \frac{1.000 g}{24.305 g/mol} = 0.041144 mol[/tex]
Let's say that x mol react in the first reaction and y mol react in the second reaction. This means:
[tex]x + y = 0.041144 mol[/tex]
According to stoichiometry, we form:
[tex]n_{MgO} = x mol, n_{Mg_3N_2} = \frac{y}{3} mol[/tex]
Multiplying moles by the molar mass of each substance will yield mass. This means we form a total of:
[tex]m_{MgO} = 40.31x g, m_{Mg_3N_2} = \frac{y}{3} 100.95 g = [/tex]
The total mass is given, so we have our second equation to solve:
[tex]40.31x + 33.65y = 1.584[/tex]
We have two unknowns and two equations, we may then solve:
[tex]x + y = 0.041144[/tex]
[tex]40.31x + 33.65y = 1.584[/tex]
Express y from the first equation:
[tex]y = 0.041144 - x[/tex]
Substitute into the second equation:
[tex]40.31x + 33.65(0.04144 - x) = 1.584[/tex]
[tex]40.31x + 1.39446 - 33.65x = 1.584[/tex]
[tex]6.66x = 0.18954[/tex]
[tex]x = 0.028459[/tex]
[tex]y = 0.041144 - x = 0.012685[/tex]
Moles of nitride formed:
[tex]n_{Mg_3N_2} = \frac{y}{3} = 0.0042282 mol[/tex]
Convert this to mass:
[tex]m_{Mg_3N_2} = 0.0042282 mol\cdot 100.95 g/mol = 0.4268 g[/tex]
Find the percentage:
[tex]\omega_{Mg_3N_2} = \frac{0.4268 g}{1.584 g}\cdot 100\% = 26.95 \%[/tex]
