com OWLV2 1 Online teaching and learning resource Use the References to access i

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com OWLV2 1 Online teaching and learning resource Use the References to access important values if needed for this question. For the following reaction, 11.1 grams of carbon monoxide are allowed to react with 11.4 grams of water carbon monoxide (g)+ water (1)carbon dioxide (g)+ hydrogen (g) What is the maximum amount of carbon dioxide that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams Submit Answer Retry Entire Group 4more group attempts remaining Cengage Learning Cengnge Technical

Explanation / Answer

Carbon monoxide (g) + Water (l) --------> Carbon Dioxide (g) + Hydrogen (g)

CO (g) + H2O (l) -------> CO2 (g) + H2(g)

number of moles of CO = 11.1/28 = 0.3964 mol

number of moles of H2O = 11.4/18 = 0.6333 mol

from the reaction 1 moles of CO requires 1 mole of H2O, Hence CO is limiting reagent

1 mole of CO produces 1 mole of CO2

0.3964 moles of CO produces 0.3964 moles of CO2 will be produced

0.3964 moles of CO requires 0.3964 moles of H2O

Excess reagent = 0.6333 - 0.3964 = 0.2369 mol of H2O

mass of excess reagent = 0.2369 * 18 = 4.2642 g of H2O

2 C4H10 (g) + 13 O2 (g) -----> 8 CO2 (g) + 10 H2O (g)

number of moles of C4H10 = 9.1/58 = 0.1569 moles

number of moles of O2 = 17.2/32 = 0.5375 moles

2 moles of C4H10 produces 8 moles of CO2

0.1569 moles of C4H10 produces 8/2*0.1569 moles of O2

                              = 0.6276 moles

13 moles of O2 produces 8 moles of CO2

0.5375 moles of O2 produces 8/13*0.5375 moles of CO2

                           = 0.3308 moles

Hence O2 is limiting reagent.

Maximum amount of CO2 formed = 0.3308 * 44 = 14.5552 g

13 moles of O2 require 2 moles of Butane

0.5375 moles of O2 requires 2/13*0.5375 moles of Butane

                           = 0.08269 moles

amount of excess reagent remaining = 0.1569 - 0.08269 = 0.07421 moles

mass of Excess reagent = 0.07421 * 58 = 4.3042 g

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