Question

Magnesium and nitrogen react in a combination reaction to produce magnesium nitride:

Answer 1

Answer:

21.7 g

Explanation:

Step 1: Write the balanced equation

3 Mg + N₂ → Mg₃N₂

Step 2: Calculate the moles corresponding to 8.33 g of nitrogen

The molar mass of N₂ is 28.01 g/mol.

$8.33 g \times \frac{1mol}{28.01g} =0.297mol$

Step 3: Calculate the moles of magnesium that reacts with 0.297 moles of nitrogen

The molar ratio of Mg to N₂ is 3:1. The reacting moles of Mg are 3/1 × 0.297 mol = 0.891 mol

Step 4: Calculate the mass corresponding to 0.891 moles of magnesium

The molar mass of Mg is 24.31 g/mol.

$0.891 mol \times \frac{24.31g}{mol} = 21.7 g$

Answer 2

Answer:

$m_{Mg}=21.7 g Mg$

Explanation:

Hello,

In this case, considering the given reaction, we are able to compute the mass of magnesium that is consumed by considering its molar mass (24.31 g/mol), the molar mass of diatomic nitrogen (28.02 g/mol), the initial mass of nitrogen (8.33 g) and the 3:1 molar ratio of magnesium to nitrogen in the reaction.

Hence we compute it by applying the shown below stoichiometric procedure:

$m_{Mg}=8.33 gN_2*\frac{1molN_2}{28.02gN_2} *\frac{3molMg}{1molN_2} *\frac{24.31gMg}{1molMg} \\\\m_{Mg}=21.7 g Mg$

Regards.