Answer:
104.56 g
Explanation:
Equation of the reaction is given as:
3 Mg + N₂ ⇒ Mg₃N₂
Given that:
1.3052 moles of magnesium is heated with 0.3782 moles of nitrogen forming magnesium nitride.
the molar mass of magnesium = 24 × 3 = 72 g/mol
the molar mass of nitrogen = 14 × 2 = 28 g/mol
we can determine their corresponding mass in gram;
for magnesium;
[tex]number of moles =\frac{mass}{molar mass}[/tex]
[tex]1.3052=\frac{mass}{72}[/tex]
[tex]mass=1.3052*72[/tex]
mass(g) of magnesium = 93.97 g
for nitrogen;
[tex]number of moles =\frac{mass}{molar mass}[/tex]
[tex]0.3782= \frac{mass}{28}[/tex]
[tex]mass= 0.3782*28[/tex]
mass(g) of nitrogen = 10.59 g
∴ the total gram that is produced when 1.3052 moles of magnesium is heated with 0.3782 moles of nitrogen forming magnesium nitride will be:
= 93.97 g + 10.59 g
= 104.56 g
Answer:
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.
