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

26.95 %

Step-by-step 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) = (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) = (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) = (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) = (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) = (0.4268 g)/(1.584 g)\cdot 100\% = 26.95 \%`