Question

Gascous butane (CH(CH), CH,) reacts with gaseous oxygen gas (0,) to produce gaseous carbon dioxide (co,) and gueous water (1,0). 16.695 of water is produced from the reaction of 8.14 g of butane and 579 g of oxygen gas, calculate the percent yield of water Round your answer to 3 significant figures Is XI ?

Answer 1

Moles of butane = mass/molar mass

= 8.14/58.12 = 0.14

Moles of Oxygen = 57.9/32 = 1.81

Reaction :

C4H10 + 13/2 O2 -> 4CO2 + 5H2O

1 mole butane require 13/2 mole O2

0.14 mole butane require 0.14×13/2 mole O2

= 0.91 mole O2

Since, we have more than this O2 thus butane is limiting reagent.

1 mole butane give 5 mole Water

0.14 mole butane give 5×0.14 = 0.70 mole water

= 0.70×18 g water

= 12.6 g water

% yield = 6.69×100/12.6 = 53.1%