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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.

Respuesta :

dsdrajlin

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.

[tex]8.33 g \times \frac{1mol}{28.01g} =0.297mol[/tex]

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.

[tex]0.891 mol \times \frac{24.31g}{mol} = 21.7 g[/tex]

sebassandin

Answer:

[tex]m_{Mg}=21.7 g Mg[/tex]

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:

[tex]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[/tex]

Regards.

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