Question
How many grams of propane (C3 H8 ) will react with 3.29 L of O at 1.05 atm and -34°C?
i know the balanced eqn is 1C3H8 + 5 O2 -----> 3CO2 + 4H20
and that PV=nrt , with T= -34 + 273.15 = 239.15K.
n=(PV)/(RT) (F=for O2). = 0.176 mol O2
0.176 mol O2 * (1mol C3H8 / 5mol O2) * 44.0962gC3H8/1mol C3H8)
I get 1.552 g of C3H8 react with O2 but i feel that its incorrect and was wondering if i iwas doing it properly
i know the balanced eqn is 1C3H8 + 5 O2 -----> 3CO2 + 4H20
and that PV=nrt , with T= -34 + 273.15 = 239.15K.
n=(PV)/(RT) (F=for O2). = 0.176 mol O2
0.176 mol O2 * (1mol C3H8 / 5mol O2) * 44.0962gC3H8/1mol C3H8)
I get 1.552 g of C3H8 react with O2 but i feel that its incorrect and was wondering if i iwas doing it properly
Answers
DrBob222
Not only have you solved the problem correctly but you have arrived at the correct answer.