WA04_CHE-121-mar14 - Written Assignment 4 Chemical...

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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. Chemical reaction is when a substance is changed into one or more new substances. In this case hydrogen and oxygen combine to form water. c. Reactants are the materials on the left side, which start the reaction. d. Products are the materials on the right side, which are the result of the reaction. 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 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) (Reference: Chang 3.47) Al + I 2 → AlI 3 20.4g Al x 1 mol Al / 26.981g Al = 7.5609 x 10 -1 mol Al 1 mol Al = 1 mol I 2; 7.5609 x 10 -1 mol Al x 1 mol I 2 / 1 mol Al = 7.5609 x 10 -1 mol I 2 7.5609 x 10 -1 mol I 2 x 253.80g I 2 = 191.90g I 2 Short version 1 mol Al = 26.981g Al; 20.4g/26.981g = 7.5609 x10 -1 mol Al 1 mol I = 126.90g I x 2 = 253.8g I 2 * 7.5609 x10 -1 mol = 191.90g I 2 4.
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