For the reaction
3 Mg + N2 → Mg3N2 ,
what is the maximum amount of Mg3N2 which
could be formed from 6.66 mol of Mg and
9.34 mol of N2?
Answer in units of mol.
Slaked lime (Ca(OH)2) is formed from quicklime
(CaO) by the addition of water:
CaO(s) + H2O(ℓ) → Ca(OH)2(s).
What mass of slaked lime can be produced
from a mixture of 29.8 g of CaO and 10.3 g of
H2O?
Answer in units of g.
2.22
To calculate the maximum amount of Mg3N2 that could be formed, we need to determine the limiting reactant between Mg and N2. The limiting reactant is the reactant that is completely consumed and limits the amount of product that can be formed.
To find the limiting reactant, we can compare the moles of each reactant to their respective stoichiometric coefficients in the balanced chemical equation:
3 Mg + N2 → Mg3N2
According to the balanced equation, the stoichiometric ratio between Mg and Mg3N2 is 3:1, and the stoichiometric ratio between N2 and Mg3N2 is 1:1.
First, we calculate the amount of Mg3N2 that can be formed from 6.66 mol of Mg:
Using the stoichiometric ratio of 3:1, we can convert moles of Mg to moles of Mg3N2:
6.66 mol Mg × (1 mol Mg3N2 / 3 mol Mg) = 2.22 mol Mg3N2 from Mg
Next, we calculate the amount of Mg3N2 that can be formed from 9.34 mol of N2:
Using the stoichiometric ratio of 1:1, we can convert moles of N2 to moles of Mg3N2:
9.34 mol N2 × (1 mol Mg3N2 / 1 mol N2) = 9.34 mol Mg3N2 from N2
Now, we compare the amounts of Mg3N2 formed from each reactant. The smaller value will be the maximum amount formed because it is limited by the reactant that gives the smaller number of moles:
The maximum amount of Mg3N2 that can be formed is 2.22 mol.