Question

a) Determine the balanced chemical equation for this reaction. C 8 H 18 (g)+ O 2 (g)→C O 2 (g)+ H 2 O(g) Enter the coefficients for each compound in order, separated by commas. For example, 1,2,3,4 would indicate one mole of C 8 H 18 , two moles of O 2 , three moles of C O 2 , and four moles of H 2 O . b) 0.370 mol of octane is allowed to react with 0.650 mol of oxygen. Which is the limiting reactant? c) How many moles of water are produced in this reaction? d) After the reaction, how much octane is left?

Answer 1

a)

The reaction is:

C8H18(g) + O2(g) —> CO2(g) + H2O(g)

Balance C:

C8H18(g) + O2(g) —> 8CO2(g) + H2O(g)

Balance H:

C8H18(g) + O2(g) —> 8CO2(g) + 9 H2O(g)

Balance O:

2 C8H18(g) + 25 O2(g) —> 16 CO2(g) + 18 H2O(g)

Answer: 2, 25, 16, 18

b)

Balanced chemical equation is:

2 C8H18 + 25 O2 ---> 18 H2O + 16 CO2

2 mol of C8H18 reacts with 25 mol of O2

for 0.37 mol of C8H18, 4.625 mol of O2 is required

But we have 0.65 mol of O2

so, O2 is limiting reagent

Answer: O2

c)

According to balanced equation

mol of H2O formed = (18/25)* moles of O2

= (18/25)*0.65

= 0.468 mol

Answer: 0.468 mol

d)

According to balanced equation

mol of C8H18 reacted = (2/25)* moles of O2

= (2/25)*0.65

= 5.2*10^-2 mol

mol of C8H18 remaining = mol initially present - mol reacted

mol of C8H18 remaining = 0.37 - 5.2*10^-2

mol of C8H18 remaining = 0.318 mol

Answer: 0.318 mol