Question

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

3 Mg + N2 → Mg3N2

In a particular experiment, a 8.33-g sample of N2 reacts completely. The mass of Mg consumed is ________ g.

Answer 1

`m_(Mg)=21.7 g Mg`

Step-by-step 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*(1molN_2)/(28.02gN_2) *(3molMg)/(1molN_2) *(24.31gMg)/(1molMg) \\\\m_(Mg)=21.7 g Mg`

Regards.

Answer 2

21.7 g

Step-by-step 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 * (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 * (24.31g)/(mol) = 21.7 g`