Balanced equation.
C7H16 + 11O2 >> 7CO2 + 8H2O
2.00 moles C7H16 (7moles CO2/1mole C7H16) = 14 moles CO2
( I forget STP standard, so will use PV = nRT )
(1atm)(V) = (14moles CO2)(0.08206 Latm/molK)(273.15K)
Volume CO2 = 14 *0.08206*273.15/1
= 314 Liters CO2
Heptane is C7H12. Hydrocarbons burn to form CO2 and water. If we balance the equation we get
C7H12 + 10 O2 -> 7 CO2 + 6 H2O
We see that for every mole of heptane burned we get 7 moles of CO2. Therefore 2 moles of heptane would get us 14 moles of CO2.
1 mole of any gas at standard temperature and pressure (STP) is equal to 22.4 liters. Multiply 22.4 liters by 14 moles to get 313.6 liters of carbon dioxide.
When the combustion is complete, the balanced equation is: C7H16 + 11 O2 -> 7 CO2 + 8 H2O.
C7H16+11O2 = 7CO2+8H2O is the balanced equation for the complete combustion of heptane.
9
The balanced equation for complete combustion of heptane is C7H16 + 11O2 ---(ignition)---> 7CO2 + 8H2O. For each moelcule of heptane, you would need 11 molecules of oxygen gas.
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This equation is C7H16 + 15 O2 -> 7 CO2 + 8 H2O.
When the combustion is complete, the balanced equation is: C7H16 + 11 O2 -> 7 CO2 + 8 H2O.
how many grams of oxygen are consumed when 19.4g of carbon dioxide is formed during the combustion of C7H16
C7H16+11O2 = 7CO2+8H2O is the balanced equation for the complete combustion of heptane.
9
For the complete combustion reaction, the equation is: C7H16 (l) + 11 O2 (g) => 7 CO2 (g) + 8 H2O (g).
C7H16 + 1102 ------->8H2O + 7CO2 So 1 molecule of heptane produces 8 molecules of water on combustion and thus 3 molecules produces 24 molecules of water.
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The balanced equation for complete combustion of heptane is C7H16 + 11O2 ---(ignition)---> 7CO2 + 8H2O. For each moelcule of heptane, you would need 11 molecules of oxygen gas.
the answer is D.
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Yes, 2,3-dimethylpentane has the empirical formula C7H16.