If 1.3052 moles of magnesium is heated with 0.3782 moles of nitrogen forming magnesium nitride, how many grams of magnesium nitride will be produced?

18.91 g.

Explanation:

Equation of the reaction:

3Mg + N2 --> Mg3N2

By stoichiometry, 3 moles of Magnesium reacted with 2 moles of Nitrogen to give 1 mole of Magnesium nitiride.

Determining the limiting reagent,

1.3052 mol of Mg * 2 mol lf N2/3 mol of Mg

= 0.87 mol on N2 (0.3782 mol of N2 is present)

N2 is the limiting reagent.

Since 2 moles of N2 reacted to give 1 mole of Magnesium nitride. Therefore,

Moles of Mg3N2 = 0.3782/2

= 0.1891 mol.

Molar mass of Mg3N2 = (24*3) + (14*2)

= 100 g/mol

Mass = molar mass * number of moles

= 100 * 0.1891

= 18.91 g of Mg3N2 is produced.

i believe the answer is 5,000