Question

Magnesium metal burns in air to form a mixture of magnesium oxide (MgO, M = 40.31) and magnesium nitride (Mg3N2, M = 100.95). A 1.000 g sample of magnesium ribbon is burned in air to give 1.584 g of the oxide/nitride mixture. What percentage of the magnesium is present in the form of the nitride?

Answer 1

Answer:

26.95 %

Explanation:

Air contains the highest percentage of oxygen and nitrogen gases. Magnesium then combines with both of the gases:

$2 Mg (s) + O_2 (g)\rightarrow 2 MgO (s)$

$3 Mg (s) + N_2 (g)\rightarrow Mg_3N_2 (s)$

Firstly, find the total number of moles of magnesium metal:

$n_{Mg} = \frac{1.000 g}{24.305 g/mol} = 0.041144 mol$

Let's say that x mol react in the first reaction and y mol react in the second reaction. This means:

$x + y = 0.041144 mol$

According to stoichiometry, we form:

$n_{MgO} = x mol, n_{Mg_3N_2} = \frac{y}{3} mol$

Multiplying moles by the molar mass of each substance will yield mass. This means we form a total of:

$m_{MgO} = 40.31x g, m_{Mg_3N_2} = \frac{y}{3} 100.95 g =$

The total mass is given, so we have our second equation to solve:

$40.31x + 33.65y = 1.584$

We have two unknowns and two equations, we may then solve:

$x + y = 0.041144$

$40.31x + 33.65y = 1.584$

Express y from the first equation:

$y = 0.041144 - x$

Substitute into the second equation:

$40.31x + 33.65(0.04144 - x) = 1.584$

$40.31x + 1.39446 - 33.65x = 1.584$

$6.66x = 0.18954$

$x = 0.028459$

$y = 0.041144 - x = 0.012685$

Moles of nitride formed:

$n_{Mg_3N_2} = \frac{y}{3} = 0.0042282 mol$

Convert this to mass:

$m_{Mg_3N_2} = 0.0042282 mol\cdot 100.95 g/mol = 0.4268 g$

Find the percentage:

$\omega_{Mg_3N_2} = \frac{0.4268 g}{1.584 g}\cdot 100\% = 26.95 \%$