Answer:
a) C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
b) You need 5 moles of Oâ‚‚ per mole of propane.
c) 181 g of Oâ‚‚
d) 254 L of Oâ‚‚ and 1210 L of air
e) 152 L of carbon dioxide
f) The gross heat released is 2354 kJ
Explanation:
C₃H₈ + O₂ → CO₂ + H₂O
a) To balance a combustion reaction you must add COâ‚‚ as carbons of hydrocarbons you have. Then, you should add waters as half of hydrogens of the hydrocarbon and, in the last, balance oxygen with Oâ‚‚, thus:
C₃H₈ + O₂ → 3 CO₂ + 4 H₂O
10 oxygens, so you sholud add 5 Oâ‚‚:
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
b) The equation balanced says that you need 5 moles of Oâ‚‚ per mole of propane.
c) To burn 100g of propane you need:
100 g C₃H₈×[tex]\frac{1 mol}{44,1g}[/tex]×[tex]\frac{5 molO_2}{1mol C_{3}H_{8}}[/tex]×[tex]\frac{16g}{1mol O_2}[/tex]= 181 g of O₂
d) 181g of Oâ‚‚ are 11,34 moles. The volume you require is:
V =nRT/P
where:
n are moles (11,34 moles)
R is gas constant (0,082 atmL/molK)
T is temperature (273 K at STP)
P is pressure (1 atm at STP)
V is 254 L of oxygen.
The liters of air are:
254L Oâ‚‚ â‚“ [tex]\frac{100 air}{21 O_2}[/tex] = 1210 L of air
e) The volume of COâ‚‚ produced is:
100 g C₃H₈×[tex]\frac{1 mol}{44,1g}[/tex]×[tex]\frac{3 molCO_2}{1mol C_{3}H_{8}}[/tex]= 6,80 moles of CO₂
V =nRT/P
where:
n are moles (6,80 moles)
R is gas constant (0,082 atmL/molK)
T is temperature (273 K at STP)
P is pressure (1 atm at STP)
V is 152 L of carbon dioxide.
f) C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O ΔH = -103,8 kJ/mol
1 kg of C₃H₈ are:
1000 g × [tex]\frac{1mol}{44,1 g}[/tex] = 22,68 moles
Thus, the gross heat released is:
103,8 kJ/mol × 22,68 moles = 2354 kJ
I hope it helps!
Answer:
propane C3H8 +O2 =CO2 +H2O
Explanation:
C3H8 + 5O2 = 3CO2 + 4H2O
3C & 10 O Â & 8 H on each side
