Answer
a. There are 10.4 g of $PbCl_2$
b. 18.8 mL of $2.00 \space M \space Pb(NO_3)_2$ will react with this other solution.
c. $1.20 \space M$ $KCl$ solution.
Work Step by Step
a.
1. Find the conversion factors
- 1.50 M $KCl$ solution: $$1.50 \space moles \space KCl = 1 \space L \space solution$$
- 1 L = 1000 mL
- $ 1 \space mole \space PbCl_2 = 2 \space moles \space KCl $
$ PbCl_2 $ : ( 35.45 $\times$ 2 )+ ( 207.2 $\times$ 1 )= 278.1 g/mol
$$ \frac{1 \space mole \space PbCl_2 }{ 278.1 \space g \space PbCl_2 } \space and \space \frac{ 278.1 \space g \space PbCl_2 }{1 \space mole \space PbCl_2 }$$
2. Calculate the mass in g of $PbCl_2$:
$$ 50.0 \space mL \space KCl \space solution \times \frac{1 \space L}{1000 \space mL} \times \frac{1.50 \space moles \space KCl }{1 \space L \space KCl \space solution} \times \frac{ 1 \space mole \space PbCl_2 }{ 2 \space moles \space KCl } \times \frac{ 278.1 \space g \space PbCl_2 }{1 \space mole \space PbCl_2 } = 10.4 \space g \space PbCl_2 $$
b.
1. Find the conversion factors
- 1 L = 1000 mL
- 1.50 M $KCl$ solution: $$1.50 \space moles \space KCl = 1 \space L \space solution$$
- $ 1 \space mole \space Pb(NO_3)_2 = 2 \space moles \space KCl $
- 2.00 M $KCl$ solution: $$2.00 \space moles \space KCl = 1 \space L \space solution$$
2. Calculate the volume of the other solution:
$$ 50.0 \space mL \space KCl \space solution \times\frac{1 \space L}{1000 \space mL} \times \frac{1.50 \space moles \space KCl }{1 \space L \space KCl \space solution} \times \frac{ 1 \space mole \space Pb(NO_3)_2 }{ 2 \space moles \space KCl } \times \frac{1 \space L \space Pb(NO_3)_2 \space solution}{2.00 \space moles \space Pb(NO_3)_2}\times \frac{1000 \space mL}{1 \space L} = 18.8 \space mL \space Pb(NO_3)_2 \space solution$$
c.
1. Find the conversion factors
- 1 L = 1000 mL
- 0.400 M $Pb(NO_3)_2$ solution: $$0.400 \space mole \space Pb(NO_3)_2 = 1 \space L \space solution$$
- $ 2 \space moles \space KCl = 1 \space mole \space Pb(NO_3)_2 $
$ KCl $ : ( 35.45 $\times$ 1 )+ ( 39.10 $\times$ 1 )= 74.55 g/mol
$$ \frac{1 \space mole \space KCl }{ 74.55 \space g \space KCl } \space and \space \frac{ 74.55 \space g \space KCl }{1 \space mole \space KCl }$$
2. Calculate the amount in moles of $KCl$:
$$ 30.0 \space mL \space Pb(NO_3)_2 \space solution \times \frac{1 \space L}{1000 \space mL} \times \frac{0.400 \space mole \space Pb(NO_3)_2 }{1 \space L \space Pb(NO_3)_2 \space solution} \times \frac{ 2 \space moles \space KCl }{ 1 \space mole \space Pb(NO_3)_2 } = 0.0240 \space mole \space KCl $$
3. Calculate the molarity:
$$Volume(L) = 20.0 \space mL \times \frac{1 \space L}{1000 \space mL} = 0.0200 \space L $$
$$Molarity(M) = \frac{0.0240 \space mole\space KCl}{0.0200 \space L} = 1.20 \space M \space KCl$$
