The following is a Limiting Reactant problem: Magnesium nitride is formed in the reaction of magnesium metal with nitrogen gas in this reaction: 3 Mg(s) + N2(g) -> Mg3N2(s)
How many grams of product are formed from 2.0 mol of N2 (g) and 8.0 mol of Mg(s)?
Do you have any choices or is it one where you answer it with what you know?
Mole Ratio of Mg : Mg3N2 = 3 b: 1
mole of Mg3N2 = moleofmagnesium over 3 = 8.0mol over 3 = 2.67 mol
mass of Mg3N2 = mole * molar mass = 2.67 mol * ((3* 24) + (2 * 14)) g / mol = 2.66.67
I hope that helped you out :)
To solve this limiting reactant problem, you need to determine which reactant will be completely consumed, thus limiting the extent of the reaction. The reactant that is completely consumed is called the limiting reactant.
To find the limiting reactant, you can compare the moles of each reactant with the stoichiometric coefficients in the balanced equation.
Given:
- Moles of N2 (g) = 2.0 mol
- Moles of Mg (s) = 8.0 mol
Based on the balanced equation: 3 Mg(s) + N2(g) -> Mg3N2(s)
You can convert the moles of each reactant to moles of the product magnesium nitride (Mg3N2) using the stoichiometric coefficients.
For N2 (g):
By looking at the balanced equation, you can see that it takes 1 mol of N2 to produce 1 mol of Mg3N2.
Therefore, with 2.0 mol of N2, you would produce 2.0 mol of Mg3N2.
For Mg (s):
From the balanced equation, you can see that it takes 3 moles of Mg to produce 1 mol of Mg3N2.
So, 8.0 mol of Mg would produce 8.0/3 = 2.67 mol of Mg3N2.
Now, you compare the moles of product obtained from each reactant.
- N2 gives 2.0 mol of Mg3N2.
- Mg gives 2.67 mol of Mg3N2.
Since N2 gives a lower amount of product (2.0 mol), N2 is the limiting reactant.
To calculate the mass of product formed, use the molar mass of Mg3N2, which is 100.93 g/mol.
Multiply the moles of Mg3N2 produced by the molar mass to get the mass:
Mass of Mg3N2 = 2.0 mol × 100.93 g/mol = 201.86 g
Therefore, 201.86 grams of product (Mg3N2) are formed from 2.0 mol of N2 (g) and 8.0 mol of Mg (s).