Practice_EXAM_II

984atmwecancalculatetheactualyieldofcarbondioxideinmol

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Unformatted text preview: ula unit (see Example 3.11 in Chapter 3) and thus the molecular formula is identical to the empirical formula as SF4. 15. The combustion process for methane, major component of natural gas, is CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) If 15.0 moles of methane are reacted, what is the volume of carbon dioxide (in L) produced at 23.0 oC and 0.985 atm? (a) 370 (b) 430 (c) 510 (d) 630 (e) 720 Hint: Gas stoichiometry: 11th ed. p.p. 193‐195 Examples 5.11 and 5.12; 10th ed. p.p. 194‐196 Examples 5.12 & 5.13; 9th ed. p.p. 190‐192. Example 5.11. This question is taken from p. 211 (5.52). From 15.0 mole of methane, we can calculate the theoretical yield in mole of carbon dioxide. Then apply PV = nRT to calculate volume in liter for carbon dioxide. Here, mole of carbonxide = mole of methane = 15 moles and thus V = nRT/P = 15x0.082x(23+273.15)/0.985 = 369.81 L. 16. In alcohol fermentation, yeast converts glucose to ethanol and carbon dioxide: C6H12O6(s) 2 C2H5OH(l)) + 2 CO2(g) If 5.97 g of glucose are reacted and 1.44 L of carbon dioxide gas are collected at 293 K and 0.984 atm, what is the percent yield of the reaction? (a) 88.9 % (b) 76.3% (c) 65.9% (d) 56.2% (e) 47.6% Hint: Gas stoichiometry: 11th ed. p.p. 193...
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This note was uploaded on 01/25/2014 for the course CHEM 1411 taught by Professor None during the Fall '11 term at HCCS.

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