Answer
See the answer below.
Work Step by Step
A) $C_3H_8(g)+5\ O_2(g)\rightarrow 3\ CO_2(g)+4\ H_2O(l)$
B) Number of moles:
$n=0.1\ atm×1.50\ L/((0.082)(293\ K))=0.00624\ mol$
C) From Dalton's law:
$0.1\ atm/(0.10 +5.00\ atm)×100\%=1.96\%$
D) Number of moles of excess oxygen:
$n=(5.0-5×0.1)\ atm×1.50\ L/((0.082)(293\ K))$
$n=0.281\ mol$
E) $P=3×0.00624\ mol×0.082×316.2\ K/1.50\ L$
$P=0.324\ atm$
F) $P=0.281\ mol×0.082×316.2\ K/1.50\ L$
$P=4.86\ atm$
