How many (a) calcium ions and how many (b) nitride ions are there in 1.00 mol Ca3N2?
a
To determine the number of calcium ions (Ca2+) and nitride ions (N3-) in 1.00 mol of Ca3N2, we need to use the ratio between the formula units and the ions.
The chemical formula Ca3N2 tells us that there are three calcium ions for every two nitride ions in the compound.
(a) Calcium ions (Ca2+):
Using the given ratio, we can calculate the number of calcium ions:
1.00 mol Ca3N2 x (3 mol Ca2+ / 1 mol Ca3N2) = 3.00 mol Ca2+
Therefore, there are 3.00 mol of calcium ions in 1.00 mol of Ca3N2.
(b) Nitride ions (N3-):
Using the given ratio, we can calculate the number of nitride ions:
1.00 mol Ca3N2 x (2 mol N3- / 1 mol Ca3N2) = 2.00 mol N3-
Therefore, there are 2.00 mol of nitride ions in 1.00 mol of Ca3N2.
To determine the number of calcium ions (Ca2+) and nitride ions (N3-) in 1.00 mol of Ca3N2, we need to understand the chemical formula of the compound.
The subscript "3" in Ca3N2 indicates that there are three calcium ions (Ca2+) in the compound. Each calcium ion has a charge of +2, denoted by the superscript "2+".
The subscript "2" in Ca3N2 indicates that there are two nitride ions (N3-) in the compound. Each nitride ion has a charge of -3, denoted by the superscript "3-".
Now, since we have 1.00 mol of Ca3N2, we need to use Avogadro's number (6.022 x 10^23) to convert the moles to number of ions.
For calcium ions (Ca2+):
Number of calcium ions = number of moles of Ca3N2 × Avogadro's number × number of calcium ions per formula unit
= 1.00 mol × 6.022 x 10^23 mol^-1 × 3
For nitride ions (N3-):
Number of nitride ions = number of moles of Ca3N2 × Avogadro's number × number of nitride ions per formula unit
= 1.00 mol × 6.022 x 10^23 mol^-1 × 2
Calculating these values, we get:
Number of calcium ions = 1.00 mol × 6.022 x 10^23 mol^-1 × 3 = 1.807 x 10^24 calcium ions
Number of nitride ions = 1.00 mol × 6.022 x 10^23 mol^-1 × 2 = 1.204 x 10^24 nitride ions
Therefore, in 1.00 mol of Ca3N2, there are approximately 1.807 x 10^24 calcium ions and 1.204 x 10^24 nitride ions.
a. 3 moles of Ca ions
b. 2 moles of N ions