WA 04_CHE-121-jan18.docx - Name College ID Thomas Edison...

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Name: College ID: Thomas Edison State University General Chemistry I with Labs (CHE-121) Section no.: Semester and year: Written Assignment 4: Chemical Reactions Answer all assigned questions and problems, and show all work. 1. Write the (a) balanced equation for the formation of liquid water from hydrogen and oxygen gas, and use it to explain the following terms: (b) chemical reaction, (c) reactant, (d) product. (12 points) a) 2H 2 + O 2 → 2H 2 O b) A chemical reaction is the changing of one or more substances into a different substance or substances. In the reaction above, two substances (hydrogen and oxygen) are combined and changed into a third substance, water. c) Reactants in chemical reactions are the substances which are changed into something else. In a chemical reaction equation, reactants are listed on the left side of the equation; in the equation above, hydrogen and oxygen are the reactants. d) Products in a chemical reaction are what is created by the combination of two reactants. In a chemical reaction equation, the product is listed on the right of the equation, after the arrow. In the equation above, water is the product. 2. Balance the following equations: (18 points) a. C + O 2 CO 2C + O 2 → 2CO b. CO + O 2 CO 2 2CO + O 2 → 2CO 2 c. Na + H 2 O H 2 + NaOH 4Na + 4H 2 O → 2H 2 + 4NaOH d. Zn + HCl ZnCl 2 + H 2 2Zn + 4HCL → 2ZnCl 2 + 2H 2 e. NaOH + H 2 SO 4 H 2 O + Na 2 SO 4 2NaOH + H 2 SO 4 → 2H 2 O + Na 2 SO 4 f. NH 3 + CuO Cu + N 2 + H 2 O 2NH 3 + 3CuO → 3Cu + N 2 + H 2 O 1 Copyright © 2017 by Thomas Edison State University . All rights reserved.
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3. Calculate the mass in grams of iodine (I 2 ) that will react completely with 20.4 g of aluminum (Al) to form aluminum iodide (AlI 3 ). (5 points) 2Al + 3I 2 → 2AlI 3 1 mol Al = 26.982g Al 26.982g Al x 2 = 53.964g Al 1 mol I = 126.90g I 126.90g I x 5 = 634.5g I 53.964g Al + 634.5g I = 688.464g 2AlI 3 = 1 mol 2AlI 3 20.4g Al x (1 mol Al/26.982g Al ) = 0.76 mol Al 0.76 mol Al x (5 mol I/2 mol Al ) = 1.9 mol I needed 1.9 mol I needed x (126.90g/mol I ) = 241.11 g I needed to react completely (Reference: Chang 3.47) 4.
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