Question

Gaseous ethane (CH,CH) will react with gaseous oxygen (O.) to produce gaseous carbon dioxide (CO) and gaseous water (H,0). Suppose 29.g of ethane is mixed with 80.6 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. X 5 ?

Answer 1

the balnced equation is as follows

2 C2H6 + 7 O2 -----------------------> 4 CO2 + 6 H2O

no of moles = mass / molar mass

for C2h6 = 29g / 30.07 g/mol => 0.9644 moles

for O2 = 80.6g / 32 g/mol => 2.51875 moles

=> 2 mol C2H6 ----------------------> 7 mol O2

0.9644 mol -------------------------->?

=> 0.9644 * 7 / 2

=> 3.375 moles of O2 required but not present

=> 7 mol O2 ----------------------> 2 mol C2H6

2.51875 mol ---------------------->?

=> 2.51875 * 2 / 7

=> 0.71964 moles required

limiting reactant is O2

excess reactant is C2H6

excess reactant moles => 0.9644 - 0.71964 => 0.24477 moles

mass of ethane left over => 0.2477 mol * 30.07 g/mol => 7.36 grams

answer => 7.4 g