Respuesta :
Answer:
a) C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
b) You need 5 moles of O₂ per mole of propane.
c) 181 g of O₂
d) 254 L of O₂ 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 CO₂ 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 O₂, thus:
C₃H₈ + O₂ → 3 CO₂ + 4 H₂O
10 oxygens, so you sholud add 5 O₂:
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
b) The equation balanced says that you need 5 moles of O₂ 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 O₂ 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 O₂ ₓ [tex]\frac{100 air}{21 O_2}[/tex] = 1210 L of air
e) The volume of CO₂ 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 & 8 H on each side
