Scan_Doc0105 - 1;) n = -= = 1.75 mol 0, RT (0.082061; ....

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5.4 Gas Stoichiometry 197 Solution Where are we going? To determine the volume of CO 2 produced What do we know? CH 4 O 2 CO 2 1.65 atm 2.80 L 25°C + 273 = 298 K 1.25 atm 35.0 L 31°C + 273 = 304 K 2.50 atm ? 125°C + 273 = 398 K p V T What information do we need? .I Balanced chemical equation for the reaction .I Ideal gas law PV = nRT .I R = 0.08206 L . atm/K . mol How do we get there? We need to use the strategy for solving stoichiometry problems that we learned in Chapter 3. 1 ~ What is the balanced equation? From the description of the reaction, the unbalanced equation is CH 4 (g) + 02(g) ------+ CO 2 (g) + H 2 0(g) which can be balanced to give CH 4 (g) + 20 2 (g) ------+ COig) + 2H 2 0(g) 2 ~ What is the limiting reactant? We can determine this by using the ideal gas law to determine the moles for each reactant: PV (1.65 atm)(2.80 1;) n = - = = 0.189 mol CH 4 RT (0.082061;· atmIR:· mol)(298 R:) . PV (1.25 atm)(35.0
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Unformatted text preview: 1;) n = -= = 1.75 mol 0, RT (0.082061; . atmIR: moI)(304 R:) In the balanced equation for the combustion reaction, 1 mol CH 4 requires 2 mol O 2 . Thus the moles of O 2 required by 0.189 mol CH 4 can be calculated as follows: 2 mol O 2 0.189 m. .el-efI 4 X = 0.378 mol O 2 1 JUGl.-H4 The limiting reactant is CH 4 . 3 ~ What are the moles of CO 2 ? Since CH 4 is limiting, we use the moles of CH 4 to determine the moles of CO 2 produced: 1 mol CO 2 0.189 JUGl.-H4 X = 0.189 mol CO 2 1 JUGl.-H4 4 ~ What is the volume of CO 2 produced? Since the conditions stated are not STP, we must use the ideal gas law to calculate the volume: V= nRT P...
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