Question

Use the balanced equation for the combustion of ethane to complete the table.

2C2H6(g)  + 7O2(g) → 4CO2(g)  + 6H2O(g)

Initially mixed	0.231 g	1.056 g	0.00 g	0.00 g
How much reacts	g	g	―	―
Composition of final mixture	g	g	g	g

Answer 1

Use the balanced equation for the combustion of ethane to complete the table.

2C2H6(g)  + 7O2(g) → 4CO2(g)  + 6H2O(g)

Initially mixed	0.231 g	1.056 g	0.00 g	0.00 g
How much reacts	0.231g	0.4298g	―	―
Composition of final mixture	0.0 g	0.6255g	0.676 g	0.415 g

Number of moles of C2H6 = 0.231 g /30.07 g/mol=7.68*10^-3 mol

Number of moles of O2= 1.056 g /15.99 g/mol=0.066 mol

Here C2H6 is limiting agent which determine the moles of product and completely use in reaction.

Mole of O2 which use to reacts with this mole of C2H6

7.68*10^-3 mol C2H6* 7 mol O2/ 2 mol C2H6=

0.02688 mol reacted

Amount of O2 = 0.02688 mol* 15.99 g/ mol=0.4298 g

Un-reacted mole = 0.066 mol-0.02688 mol=0.03912 mol

Amount of O2= 0.03912*15.99=0.6255 g

Mole of H2O:

7.68*10^-3 mol C2H6* 6 mol H2O/ 2 mol C2H6=

0.02304 mol H2O

Amount of H2O

0.02304 mol H2O *18.02 g/mol

=0.415 g

Mole of CO2:

7.68*10^-3 mol C2H6* 4 mol CO2/ 2 mol C2H6=

0.01536 mol H2O

Amount of CO2

0.01536 mol CO2 *44.00 g/mol

=0.676 g