Answer:
1) 0.428 mol
Here the compound that we have is [tex]NaCl[/tex].
We know that:
- Molar mass of sodium ([tex]Na[/tex]): [tex]22.990 g/mol[/tex]
- Molar mass of chlorine ([tex]Cl[/tex]): [tex]35.453 g/mol[/tex]
Since the ratio between [tex]Na,Cl[/tex] in the compound is 1:1, the molar mass of the [tex]NaCl[/tex] is:
[tex]M_{NaCl}=M_{Na}+M_{Cl}=22.990+35.354=58.344 g/mol[/tex]
Here the mass of the compound is
[tex]m=25 g[/tex]
Therefore, the number of moles is:
[tex]n=\frac{m}{M}=\frac{25}{58.344}=0.428 mol[/tex]
2) 1.27 mol
The compound in this problem is
[tex]H_2 SO_4[/tex]
The molar mass of each element is:
[tex]M_H=1.0079 g/mol[/tex] (hydrogen)
[tex]M_S=32.065 g/mol[/tex] (solphur)
[tex]M_O=15.994 g/mol[/tex] (oxygen)
In this compound, we have 2 atoms of hydrogen, 1 atom of solphur and 4 atoms of oxygen, therefore the total molar mass is:
[tex]M_{H_2SO_4}=2M_H+M_S+4M_O=2(1.0079)+(32.065)+4(15.994)=98.057 g/mol[/tex]
Here the mass of the compound is
[tex]m=125 g[/tex]
So the number of moles is
[tex]n=\frac{m}{M}=\frac{125}{98.057}=1.27 mol[/tex]
3) 0.633 mol
The compound in this problem is
[tex]KMnO_4[/tex]
The molar mass of each element is:
[tex]M_K=39.0983 g/mol[/tex] (potassium)
[tex]M_{Mn}=54.9380 g/mol[/tex] (manganese)
[tex]M_O=15.994 g/mol[/tex] (oxygen)
Here we have 1 atom of potassium, 1 atom of manganese and 4 atoms of oxygen, so the total molar mass of the compound is
[tex]M=M_K+M_{Mn}+4M_O=(39.0983)+(54.9380)+4(15.994)=158.012 g/mol[/tex]
Here the mass of the compound is
[tex]m=100 g[/tex]
So the number of moles is
[tex]n=\frac{m}{M}=\frac{100}{158.012}=0.633 mol[/tex]
4) 0.993 mol
The compound in this problem is
[tex]KCl[/tex]
The molar mass of each element is:
[tex]M_K=39.0983 g/mol[/tex] (potassium)
[tex]M_{Cl}[/tex] = [tex]35.453 g/mol[/tex] (chlorine)
So, the total molar mass of the compound is:
[tex]M_{KCl}=M_K+M_{Cl}=39.0983+35.453=74.551 g/mol[/tex]
The mass of the compound here is
[tex]m=74 g[/tex]
Therefore, the number of moles is
[tex]n=\frac{m}{M}=\frac{74}{74.551}=0.993 mol[/tex]
5) 0.140 mol
The compound in this problem is
[tex]CuSO_4\cdot 5H_2O[/tex]
The molar mass of each element is:
[tex]M_{Cu}=63.546 g/mol[/tex] (copper)
[tex]M_S=32.065 g/mol[/tex] (solphur)
[tex]M_O=15.994 g/mol[/tex] (oxygen)
[tex]M_H=1.0079 g/mol[/tex] (hydrogen)
Here we have 1 atom of copper, 1 of solphure, (4+5=9) of oxygen, and (5*2=10) atoms of hydrogen, so the total molar mass is
[tex]M=M_{Cu}+M_S+9M_O+10M_H=\\=63.546+32.065+9(15.994)+10(1.0079)=249.636 g/mol[/tex]
Here the mass of the compound is
[tex]m=35 g[/tex]
So, the number of moles is
[tex]n=\frac{m}{M}=\frac{35}{249.636}=0.140 mol[/tex]
