Question

PHEASE HELP ASAP

Given the following equation C3H8 + 5O2 = 3CO2 +4H2O if i perform this reaction with 3.6 moles of C3H8 and an excess of oxygen gas , what is my theoretical yeild of water in moles? If i actually isolate 12 moles of waterwhat is my percent yeild?

Answer 1

Given the balanced chemical equation for combustion of propane:

`C_(3)H_(8)+5O_(2)-->3CO_(2)+4H_(2)O`

Moles of propane
`C_(3)H_(8)`=3.6 mol

Propane is the limiting reactant as the other reactant oxygen is said to be present in excess.

Amount of products formed would depend on the moles of limiting reactant.

Mole ratio of water to propane as per the balanced chemical equation

=
`(4molH_(2)O )/(1molC_(3)H_(8))`

Calculating the moles of water produced from 3.6 mol propane:

`3.6molC_(3)H_(8)*`
`(4molH_(2)O )/(1molC_(3)H_(8))`

= 14.4mol
`H_(2)O`

Theoretical yield of water = 14.4 mol

Actual yield = 12 mol

Percent yield =
`(ActualYield)/(TheoreticalYield)*100`

=
`(12 mol)/(14.4 mol)*100=83.3%`