Answer
a. 1600 g of $Al(NO_3)_3$
b. 6.8 g $C_6H_{12}O_6$
c. 17.9 g of $LiCl$
Work Step by Step
a.
1. Find the conversion factors
- 3.0 M $Al(NO_3)_3$ solution: $$3.0 \space moles \space Al(NO_3)_3 = 1 \space L \space solution$$
- Molar mass: $ Al(NO_3)_3 $ : ( 26.98 $\times$ 1 )+ ( 14.01 $\times$ 3 )+ ( 16.00 $\times 3 \times 3$ )= 213.01 g/mol
2. Calculate the mass in grams:
$$2.5 \space L \space solution \times \frac{3.0 \space moles \space Al(NO_3)_3}{1 \space L \space solution} \times \frac{213.01 \space g \space Al(NO_3)_3}{1 \space mole \space Al(NO_3)_3} = 1600 \space g \space Al(NO_3)_3$$
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b.
1. Find the conversion factors
- 0.50 M $C_6H_{12}O_6$ solution: $$0.50 \space mole \space C_6H_{12}O_6 = 1 \space L \space solution$$
- Molar mass: $ C_6H_{12}O_6 $ : ( 1.008 $\times$ 12 )+ ( 12.01 $\times$ 6 )+ ( 16.00 $\times$ 6 )= 180.16 g/mol
- 1 L = 1000 mL
2. Calculate the mass in grams:
$$75 \space mL \space solution \times \frac{1 \space L}{1000 \space mL} \times \frac{0.50 \space mole \space C_6H_{12}O_6}{1 \space L \space solution} \times \frac{180.16 \space g \space C_6H_{12}O_6}{1 \space mole \space C_6H_{12}O_6} = 6.8 \space g \space C_6H_{12}O_6$$
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c.
1. Find the conversion factors
- 1.80 M $LiCl$ solution: $$1.80 \space moles \space LiCl = 1 \space L \space solution$$
- Molar mass: $ LiCl $ : ( 6.941 $\times$ 1 )+ ( 35.45 $\times$ 1 )= 42.39 g/mol
- 1 L = 1000 mL
2. Calculate the mass in grams:
$$235 \space mL \space solution \times \frac{1 \space L}{1000 \space mL} \times \frac{1.80 \space moles \space LiCl}{1 \space L \space solution} \times \frac{42.39 \space g \space LiCl}{1 \space mole \space LiCl} = 17.9 \space g \space LiCl$$
