Answer
a. There are necessary 120. g of NaOH.
b. There are necessary 59.6 g of KCl
c. There are necessary 5.47 g of HCl
Work Step by Step
a.
1. Find the conversion factors
- 1.50 M $NaOH$ solution: $$1.50 \space moles \space NaOH = 1 \space L \space solution$$
- Molar mass: $ NaOH $ : ( 22.99 $\times$ 1 )+ ( 1.008 $\times$ 1 )+ ( 16.00 $\times$ 1 )= 40.00 g/mol
2. Calculate the mass in grams:
$$2.00 \space L \space solution \times \frac{1.50 \space moles \space NaOH}{1 \space L \space solution} \times \frac{40.00 \space g \space NaOH}{1 \space mole \space NaOH} = 120. \space g \space NaOH$$
b.
1. Find the conversion factors
- 0.200 M $KCl$ solution: $$0.200 \space mole \space KCl = 1 \space L \space solution$$
- Molar mass: $ KCl $ : ( 35.45 $\times$ 1 )+ ( 39.10 $\times$ 1 )= 74.55 g/mol
2. Calculate the mass in grams:
$$4.00 \space L \space solution \times \frac{0.200 \space mole \space KCl}{1 \space L \space solution} \times \frac{74.55 \space g \space KCl}{1 \space mole \space KCl} = 59.6 \space g \space KCl$$
c.
1. Find the conversion factors
- 6.00 M $HCl$ solution: $$6.00 \space moles \space HCl = 1 \space L \space solution$$
- Molar mass: $ HCl $ : ( 35.45 $\times$ 1 )+ ( 1.008 $\times$ 1 )= 36.46 g/mol
- 1 L = 1000 mL
2. Calculate the mass in grams:
$$25.0 \space mL \space solution \times \frac{1 \space L}{1000 \space mL} \times \frac{6.00 \space moles \space HCl}{1 \space L \space solution} \times \frac{36.46 \space g \space HCl}{1 \space mole \space HCl} = 5.47 \space g \space HCl$$