For the reaction
? Li+? N2 →? Li3N ,
what is the maximum amount of Li3N which
could be formed from 11.79 g of Li and 17.08 g of N2?
I got 19.72 g as my final answer and I just want to make sure I am correct. If I am not can you please help me understand this problem?
I worked in in my head, 20 grams( forget my significant digits), so it looks like you are correct. Check it again, but it is reasonable.
Okay thank you! :)
To find the maximum amount of Li3N that can be formed, we need to use stoichiometry and the concept of limiting reactants.
First, we need to balance the equation:
4Li + N2 → 2Li3N
Now let's calculate the number of moles for each reactant using their respective molar masses.
Molar mass of Li = 6.94 g/mol
Molar mass of N2 = 28.02 g/mol
Moles of Li = 11.79 g / 6.94 g/mol = 1.696 mol
Moles of N2 = 17.08 g / 28.02 g/mol = 0.610 mol
Now, consider the balanced equation: 4Li + N2 → 2Li3N
The stoichiometry of the equation tells us that 4 moles of Li react with 1 mole of N2 to produce 2 moles of Li3N.
Comparing the moles of Li to N2, we can see that the mole ratio is 4:1. This means that if we have 1 mole of N2, we would need 4 moles of Li to react completely and form Li3N.
Since we have only 0.610 mol of N2, it means that we require (4 mol Li / 1 mol N2) * 0.610 mol N2 = 2.44 mol Li for complete reaction.
However, we only have 1.696 mol of Li. So, Li is the limiting reactant, and the maximum amount of Li3N that can be formed is determined by the amount of Li available.
Now, let's calculate the mass of Li3N formed from the limiting reactant, Li.
Molar mass of Li3N = 34.83 g/mol
Mass of Li3N = (1.696 mol Li) * (34.83 g/mol) = 59.10 g
Therefore, the correct answer for the maximum amount of Li3N that can be formed is 59.10 g, not 19.72 g.