Liquid hexane CH3CH24CH3 reacts with gaseous oxygen gas O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O . If 0.273g of water is produced from the reaction of 0.86g of hexane and 0.94g of oxygen gas, calculate the percent yield of water. Be sure your answer has the correct number of significant digits in it.
The balanced reaction of hexane and oxygen is given by
2(CH3(CH2)4CH3)+19O2 ----------- > 12CO2+14H2O
no of moles of hexane given = 0.86/86 = 0.01.
no of moles of oxygen gas = 0.94/32 = 0.029
Here hexane is the limiting reagent
so the amout of hexane required for 0.029 moles of oxygen is = 0.003 moles.
1 mole of hexane produces 7 moles of water
0.003 moles of hexane gives 7 * 0.003 moles of water
= 0.021 moles of water.= 0.378 gms
Percent yield = Actual yield / Theoritical yield * 100
Actual yield = 0.273 g
Theoritical yield = 0.378
Percent yield = 0.273/0.378 * 100
= 72.2 %
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