Answer
(a) sodium sulfide: Na$_2$S
(b) strontium chloride: SrCl$_2$
(c) aluminum oxide: Al$_2$O$_3$
(d) aluminum chloride: AlCl$_3$
Work Step by Step
(a) sodium sulfide: Na$_2$S
charge on sodium ion = 1+
charge on sulfur ion = 2-
The charges must be balanced in a compound.
2 sodium ions = 2(1+) = 2+ charge
1 sulfide ions = 1(2-) = 2- charge
(2+) + (2-) = 0
The charges balance out, so you need two atoms of sodium for every one atom of sulfur to make a compound.
(b) strontium chloride: SrCl$_2$
charge on strontium ion = 2+
charge on chloride ion = 1-
The charges must be balanced in a compound.
1 strontium ion = 1(2+) = 2+ charge
2 chloride ions = 2(1-) = 2- charge
(2+) + (2-) = 0
The charges balance out, so you need one atom of strontium for every two atoms of chlorine to make a compound.
(c) aluminum oxide: Al$_2$O$_3$
charge on aluminum ion = 3+
charge on oxide ion = 2-
The charges must be balanced in a compound.
2 aluminum ions = 2(3+) = 6+ charge
3 oxide ions = 3(2-) = 6- charge
(6+) + (6-) = 0
The charges balance out, so you need two atoms of aluminum for every three atoms of oxygen to make a compound.
(d) aluminum chloride: AlCl$_3$
charge on aluminum ion = 3+
charge on chloride ion = 1-
The charges must be balanced in a compound.
1 aluminum ion = 1(3+) = 3+ charge
3 chloride ions = 3(1-) = 3- charge
(3+) + (3-) = 0
The charges balance out, so you need one atom of aluminum for every three atoms of chlorine to make a compound.
