Gaseous ethane CH3CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O . Suppose 21. g of ethane is mixed with 25.9 g of oxygen. Calculate the minimum mass of ethane that could be left over by the chemical reaction. Round your answer to 2 significant digits.
The balanced chemical equation is
2 CH3-CH3 (g) +7 O2 (g) ------> 4 CO2 (g) + 6 H2O (g)
As per the balanced equation, 2 moles of ethane reacts with 7 moles of O2.
We need to find the limiting reagent.
Molar mass of ethane = 30.07 g/mol; molar mass of oxygen = 15.9994 g/mol.
21.0 g ethane will react with (21.0 g ethane)*(1 mole ethane/30.07 g ethane)*(7 mole O2/2 mole ethane)*(15.9994 g O2/1 mole O2) = 39.107 g O2
25.9 g O2 will react with (25.9 g O2)*(1 mole O2/15.9994 g O2)*(2 mole ethane/7 mole O2)*(30.07 g ethane/1 mole ethane) = 13.908 g ethane.
Offcourse, O2 is the limiting reagent and 25.9 g O2 will react with 13.908 g ethane.
The amount of ethane left unreacted = (21.0 – 13.908) g = 7.092 g ≈ 7.09 g (correct to 2 decimal places) (ans).
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