Question #bd124

2 Answers
Feb 4, 2018

Answer:

#57.4g# #CO_2#

Explanation:

#C_2H_5OH + 3O_2 → 2CO_2+ 3H_2O#

1 mole of #C_2H_5OH# produces 2 moles of #CO_2#

#=> 46g# #C_2H_5OH# produces #88g# #CO_2#

#=> 1g# #C_2H_5OH# produces #88/46g# #CO_2#

#=> 30g# #C_2H_5OH# produces #88/46*30g# #CO_2#

#=> 30g# #C_2H_5OH# produces #57.4g# #CO_2#

Feb 4, 2018

Answer:

Since #"O"_2# is used for burning, it's usually in excess. The fuel (here ethanol) is the limiting reactant and so the yield of carbon dioxide will only depend on that.

Explanation:

Let's start with finding moles of ethanol burnt in the reaction.

#"Moles of ethanol" = "Mass of ethanol"/"Molar mass of ethanol"#

#= ("30.0 g")/("46.07 g/mol") = "0.6512 mols"#

Since all of the #"C"# in ethanol converts to #"CO"_2#, we can use the stoichiometric relationship from the balanced equation to relate the moles of ethanol to moles of #"CO"_2#.

#"1 mol ethanol " -> " 2 moles CO"_2#

So, the number of moles of #"CO"_2# produced by #0.65# moles ethanol #= "0.6512 mol" *2 = "1.3024 moles"#

#"Mass of CO"_2 = "Moles CO"_2 * "Molar mass of CO"_2#

#= "1.3024 moles" * "44.0 g/mol" = "57.3 g"#