Answer
a. 206 mL of this HCl solution.
b. 11.2 L of $H_2$ gas.
c. 9.1 M HCl
Work Step by Step
a.
1. Find the conversion factors
- 1 L = 1000 mL
- Molar mass: $(Mg)$: 24.31 g/mole
- $ 2 \space moles \space HCl = 1 \space mole \space Mg $
- 6.00 M $HCl$ solution: $$6.00 \space moles \space HCl = 1 \space L \space solution$$
2. Calculate the volume of the other solution:
$$ 15.0 \space g \space Mg \times\frac{1 \space mole \space Mg}{24.31 \space g \space Mg} \times \frac{ 2 \space moles \space HCl }{ 1 \space mole \space Mg } \times \frac{1 \space L \space HCl \space solution}{6.00 \space moles \space HCl}\times \frac{1000 \space mL}{1 \space L} = 206 \space mL \space HCl \space solution$$
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b.
1. Find the conversion factors
- 2.00 M $HCl$ solution: $$2.00 \space moles \space HCl = 1 \space L \space solution$$
- $ 1 \space mole \space H_2 = 2 \space moles \space HCl $
- At STP: 1 mole gas = 22.4 L
2. Calculate the volume of the other solution:
$$ 0.500 \space L \space HCl \space solution \times\frac{2.00 \space moles \space HCl }{1 \space L \space HCl \space solution} \times \frac{ 1 \space mole \space H_2 }{ 2 \space moles \space HCl } \times \frac{22.4 \space L \space H_2}{1 \space mole \space H_2} = 11.2 \space L \space H_2 \space solution$$
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c.
1. Find the amount of moles of $H_2$ gas.
$PV = nRT \longrightarrow n = \frac{PV}{RT}$
$T/K = 25 + 273.15 = 298.15$
$$n = \frac{(735 \space mmHg)(5.20 \space L)}{(62.3 \space L \space mmHg \space K^{-1} \space mol^{-1})(298.15 \space K)} = 0.206 \space mole$$
2. Find the conversion factors and calculate the molarity of the HCl solution.
$1 \space L = 1000 \space mL$
$2 \space moles \space HCl = 1 \space mole \space H_2$
$$Molarity = \frac{n}{Volume} = \frac{(0.206 \space mole \space H_2) \times \frac{2 \space moles \space HCl}{1 \space mole H_2}}{45.2 \space mL \times \frac{1 \space L}{1000 \space mL}} $$ $$Molarity = 9.1 \space M \space HCl$$
