Gas StoichiometryAmount ofreactant (gramsor volume)Moles ofAmount ofMoles ofproduct (gramsor volume)reactantproductWhat is the volume of CO, produced at 37 °C and 1.00 atmwhen 5.60 g of glucose are used up in the reaction:C6H1206 (s) + 60, (g)–6CO2 (9) + 6H,0 ()g C3H1206 -mol C6H1206-mol CO2VCO2-1 metCGH,,O6180 g GgH12066 mol CO25.60 g CH1206 x1 metCH12O6= 0.187 mol CO2L•atm0.187 motx 0.0821x 310.15 KnRTV=P= 4.76 L1.00 atm24

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lómetry Amount of reactant (grams or volume) Moles of Amount of Moles of product (grams or volume) reactant product What is the volume of CO, produced at 37 °C and 1.00 atm when 5.60 g of glucose are used up in the reaction: CH1206 (s) + 6O2 (g) 6CO2 (g) + 6H,O () - g C6H1206– mol CgH1206 - mol CO2 - 1 metCH,,06 180 g Get1206 6 mol CO2 1 metC,H1206 5.60 9GH1206 x = 0.187 mol CO2 0.187 motx 0.0821 Leatm x 310.15 K nRT V= = 4.76 L 1.00 atm 24

lómetry Amount of reactant (grams or volume) Moles of Amount of Moles of product (grams or volume) reactant product What is the volume of CO, produced at 37 °C and 1.00 atm when 5.60 g of glucose are used up in the reaction: CH1206 (s) + 6O2 (g) 6CO2 (g) + 6H,O () - g C6H1206– mol CgH1206 - mol CO2 - 1 metCH,,06 180 g Get1206 6 mol CO2 1 metC,H1206 5.60 9GH1206 x = 0.187 mol CO2 0.187 motx 0.0821 Leatm x 310.15 K nRT V= = 4.76 L 1.00 atm 24

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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