4 2 2 7 2 2 7 3 mol ethanol 46068 g ethanol 3238 10 mol K Cr O 002238 g ethanol

4 2 2 7 2 2 7 3 mol ethanol 46068 g ethanol 3238 10

This preview shows page 9 - 10 out of 11 pages.

4 2 2 7 2 2 7 3 mol ethanol 46.068 g ethanol 3.238 10 mol K Cr O 0.02238 g ethanol 2 mol K Cr O 1 mol ethanol × × × = The percent ethanol by mass is: 0.02238 g 100% 10.0 g = × = % by mass ethanol 0.224% This is well above the legal limit of 0.08 percent by mass ethanol in the blood. The individual should be prosecuted for drunk driving. 4.155 Notice that nitrogen is in its highest possible oxidation state ( + 5) in nitric acid. It is reduced as it decomposes to NO 2 . 4HNO 3 ⎯⎯→ 4NO 2 + O 2 + 2H 2 O The yellow color of “old” nitric acid is caused by the production of small amounts of NO 2 which is a brown gas. This process is accelerated by light. 4.156 (a) Zn( s ) + H 2 SO 4 ( aq ) ⎯⎯→ ZnSO 4 ( aq ) + H 2 ( g ) (b) 2KClO 3 ( s ) ⎯⎯→ 2KCl( s ) + 3O 2 ( g ) (c) Na 2 CO 3 ( s ) + 2HCl( aq ) ⎯⎯→ 2NaCl( aq ) + CO 2 ( g ) + H 2 O( l ) (d) NH 4 NO 2 ( s ) heat ⎯⎯⎯→ N 2 ( g ) + 2H 2 O( g ) 4.157 Because the volume of the solution changes (increases or decreases) when the solid dissolves. CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS 125 4.158 NH 4 Cl exists as 4 NH + and Cl . To form NH 3 and HCl, a proton (H + ) is transferred from 4 NH + to Cl . Therefore, this is a Brønsted acid-base reaction. 4.159 (a) The precipitate CaSO 4 formed over Ca preventing the Ca from reacting with the sulfuric acid. (b) Aluminum is protected by a tenacious oxide layer with the composition Al 2 O 3 . (c) These metals react more readily with water. 2Na( s ) + 2H 2 O( l ) ⎯⎯→ 2NaOH( aq ) + H 2 ( g ) (d) The metal should be placed below Fe and above H. (e) Any metal above Al in the activity series will react with Al 3 + . Metals from Mg to Li will work. 4.160 (a) First Solution: 3 4 4 4 4 3 2 4 1 mol KMnO 0.8214 g KMnO 5.1974 10 mol KMnO 158.04 g KMnO 5.1974 10 mol KMnO mol solute 1.0395 10 L of soln 0.5000 L × = × × = = = × M M Second Solution: M 1 V 1 = M 2 V 2 (1.0395 × 10 2 M )(2.000 mL) = M 2 (1000 mL) M 2 = 2.079 × 10 5 M Third Solution: M 1 V 1 = M 2 V 2 (2.079 × 10 5 M )(10.00 mL) = M 2 (250.0 mL) M 2 = 8.316 × 10 7 M (b) From the molarity and volume of the final solution, we can calculate the moles of KMnO 4 . Then, the mass can be calculated from the moles of KMnO 4 . 7 7 4 4 7 4 4 4 8.316 10 mol KMnO 250 mL 2.079 10 mol KMnO 1000 mL of soln 158.04 g KMnO 2.079 10 mol KMnO 1 mol KMnO × × = × × × = 5 4 3.286 10 g KMnO × This mass is too small to directly weigh accurately. 4.161 (a) The balanced equations are: 1) Cu( s ) + 4HNO 3 ( aq ) ⎯⎯→ Cu(NO 3 ) 2 ( aq ) + 2NO 2 ( g ) + 2H 2 O( l ) Redox 2) Cu(NO 3 ) 2 ( aq ) + 2NaOH( aq ) ⎯⎯→ Cu(OH) 2 ( s ) + 2NaNO 3 ( aq ) Precipitation 3) Cu(OH) 2 ( s ) heat ⎯⎯⎯→ CuO( s ) + H 2 O( g ) Decomposition 4) CuO( s ) + H 2 SO 4 ( aq ) ⎯⎯→ CuSO 4 ( aq ) + H 2 O( l ) Acid-Base 5) CuSO 4 ( aq ) + Zn( s ) ⎯⎯→ Cu( s ) + ZnSO 4 ( aq ) Redox 6) Zn( s ) + 2HCl( aq ) ⎯⎯→ ZnCl 2 ( aq ) + H 2 ( g ) Redox CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS 126 (b) We start with 65.6 g Cu, which is 1 mol Cu 65.6 g Cu 1.032 mol Cu 63.55 g Cu × = . The mole ratio between product and reactant in each reaction is 1:1. Therefore, the theoretical yield in each reaction is 1.032 moles.
Image of page 9
Image of page 10

You've reached the end of your free preview.

Want to read all 11 pages?

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes