1. 12.5 g CaCl_{2}
= 0.113 mol CaCl_{2}_{
}
mol CaCl_{2}

=

12.5 g CaCl_{2}

x

1 mol CaCl_{2}

=

0.113 mol





111.1 g CaCl_{2}


CaCl_{2}

_{
}2. 10.0 g Na_{2}SO_{4}
= 0.0704 mol Na_{2}SO_{4}_{
}
mol
Na_{2}SO_{4}

=

10.0 g Na_{2}SO_{4}

x

1 mol Na_{2}SO_{4}

=

0.0704 mol





142.0 g Na_{2}SO_{4}


Na_{2}SO_{4}

3. 16 g NaOH = 0.40
mol NaOH
mol NaOH 
= 
16 g NaOH 
x 
1 mol NaOH 
= 
0.40 mol NaOH 




40.0 g NaOH 


4. 125 mg NaOH = 3.13
x 10^{3} mol NaOH
mol NaOH

=

125
mg NaOH

x

1 g

x

1
mol

=

3.13 x 10^{3} mol





1000
mg


40.0
g


NaOH

The only "twist" on this problem
is that you first must convert mg to grams. The problem is then identical
to the previous problem in terms of the approach to solving it. Notice
that the answer was given in scientific notation. Your calculator should
have read 0.003125. Remember that the answer should be reported to 3 signficant
figures so it was rounded to 0.00313 and then converted to scientific
notation because it is a small number.
5. 0.125 mol CaCl_{2}
= 13.9 g CaCl_{2}
g CaCl_{2} 
= 
0.125 mol CaCl_{2} 
x 
111.1 g CaCl_{2} 
= 
13.9 g CaCl_{2} 




1 mol CaCl_{2} 


6. 0.25 mol Na_{2}SO_{4}
= 36 g Na_{2}SO_{4}
g Na_{2}SO_{4} 
= 
0.25 mol Na_{2}SO_{4} 
x 
142.0 gNa_{2}SO_{4}

= 
36 g 




1 mol Na_{2}SO_{4} 

Na_{2}SO_{4} 
The answer before rounding was 35.5 g. However,
the answer must be reported to 2 significant figures. Therefore, it was
rounded to 36 g.
7. 1.55 mol NaOH = 62.0
g NaOH
g NaOH 
= 
1.55 mol NaOH 
x 
40.0 g NaOH 
= 
62.0 g NaOH 




1 mol NaOH 


