General, Organic, and Biological Chemistry: Structures of Life (5th Edition)

Published by Pearson
ISBN 10: 0321967461
ISBN 13: 978-0-32196-746-6

Chapter 9 - Solutions - 9.4 Solution Concentrations and Reactions - Questions and Problems - Page 348: 9.45

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: $$25.0 \space mL \space solution \times \frac{1 \space L}{1000 \space mL} \times \frac{6.00 \space moles \space HCl}{1 \space L \space solution} \times \frac{36.46 \space g \space HCl}{1 \space mole \space HCl} = 5.47 \space g \space HCl$$
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