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)?
i havent been able to figure this one out, can someone help me?
First thing first: write and balance the euqation. In this case, it is already balance. Next, given the moles of the reactants, you should be able to identify which one of them is in excess and which one is limiting agent. The ratio of Mg to N2 is 3:1, meaning if you have 2 mol of N2, you should have had 6 mol of Mg. Hence mg is in excess (containing 8 instead of 6)..
Thirdly, consider the molar ratio of the entire reaction, basing from the limiting reatant (N2). From the equation N2 to Mg3N2 are in the ratio of 1 as to 1. So The product also has 2 mol.
Fourthly, use the formula mass=mol*molecular mass. Use your period table to find the total mass of the product and multiply it with the 0.2
Got it?
sorry .-. I don't understand
This is a limiting reagent (LR) problem; you know that because amounts are given for both reactants.
3Mg(s) + N2(g) --> Mg3N2(s)
The long long way to do this.
mols Mg = 8 from the problem.
mols N2 = 2 from the problem.
Using the coefficients in the balanced equation, convert mols Mg to mols Mg3N2.
Do the same and convert mols N2 to mols Mg3N2.
It is likely that these two values will not be the same; the correct value in LR problem is ALWAYS the smaller value and the reagent responsible for that value is called the LR.
Now use the smaller value and convert to grams. g = mols Mg3N2 x molar mass Mg3N2 = ?
To find the number of grams of product formed, you need to use the given amounts of reactants and the stoichiometry of the reaction. Let's break it down step by step:
1. Determine the balanced equation:
3 Mg(s) + N2(g) → Mg3N2(s)
The stoichiometric ratio in the equation tells us that 3 moles of Mg reacts with 1 mole of N2 to produce 1 mole of Mg3N2.
2. Calculate the limiting reactant:
To determine the limiting reactant, compare the number of moles of each reactant to the stoichiometry of the balanced equation. The reactant that is completely consumed or used up first is the limiting reactant.
Given:
- 2.0 mol of N2
- 8.0 mol of Mg
According to the stoichiometry, for every 1 mole of N2, you need 3 moles of Mg. Therefore, to react with 2.0 mol of N2, you would need (3 mol Mg/1 mol N2) × (2.0 mol N2) = 6.0 mol Mg.
Since there is an excess (8.0 mol) of Mg compared to the required 6.0 mol, Mg is in excess. N2 is the limiting reactant because you only have 2.0 mol of it.
3. Calculate the moles of product:
Using the mole ratio from the balanced equation, we know that:
- For every 1 mole of N2 reacted, 1 mole of Mg3N2 is formed.
Therefore, 2.0 mol of N2 will produce 2.0 mol of Mg3N2.
4. Calculate the mass of product:
To find the mass, we need to use the molar mass of Mg3N2. The molar mass of Mg3N2 is the sum of the atomic masses of each element in the compound.
Molar mass of Mg = 24.31 g/mol
Molar mass of N = 14.01 g/mol
Molar mass of Mg3N2 = (3 × 24.31 g/mol) + (2 × 14.01 g/mol) = 100.95 g/mol
Now, we can calculate the mass of the product:
Mass of Mg3N2 = (2.0 mol N2) × (100.95 g/mol Mg3N2) = 201.9 g
Therefore, 201.9 grams of Mg3N2 are formed from 2.0 mol of N2 and 8.0 mol of Mg.