Consider the combustion of butane, C4H1O: 2 C4H10(g) + 13 O2(g) -> 8 CO2 (g) + 10 H2O(l)?

Question

In a particular reaction, 5.0 moles of butane are reacted with an excess amount of molecular oxygen. Calculate the number of moles of carbon dioxide formed. Calculate the mass of H2O formed.

Answer 1

so butane is limiter and everything scales based on butane in the BALANCED rnx



5.0 moles butane * 8 CO2/2 butane = 20 moles CO2

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5.0 moles butane * 10 moles H2O/2 moles butane * 18 g H2O/1mole H2O = 450 g H2O

Answer 2

{1} 2 moles of butane require 8 moles of carbon(iv)oxide then 5 moles of butane will require (5x8/2)moles of carbon(iv)oxide which is 20moles.

{2} 2 moles of butane require 10 moles of water then 5 moles will require (5x10/2)moles =25molesbut mass=molar mass X no. of moles=18 X 25=450g where 18= molar mass of water

Answer 3

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