yes, Pentane
Pentane is C5H12 The Structure is as follows. CH3-CH2-CH2-CH2-CH3
C5h12 + 8o2 --> 5co2 + 6h2o
To balance the combustion reaction of C5H12 + O2 → CO2 + H2O, first balance the carbon atoms, then the hydrogen atoms, and finally the oxygen atoms. In this case, the balanced equation is: C5H12 + 8 O2 → 5 CO2 + 6 H2O
Pentane
The balanced equation is: C5H12 + 8O2 → 5CO2 + 6H2O. Therefore, the coefficient for oxygen in the balanced equation is 8.
The balanced chemical equation for the combustion of pentane C5H12 is: C5H12 + 8O2 → 5CO2 + 6H2O Therefore, for every 1 mole of pentane, 8 moles of oxygen gas are required. So, 0.100 mol of pentane will require 0.100 mol * 8 = 0.800 mol of oxygen gas to react completely.
Pentane is C5H12 The Structure is as follows. CH3-CH2-CH2-CH2-CH3
pentane
C5h12 + 8o2 --> 5co2 + 6h2o
The percent composition of C5H12 is: carbon (C) = 83.78%, hydrogen (H) = 16.22%.
The molar mass of C5H12 is 72 grams/mole.
No, C5H12 is not a Lewis base. Lewis bases are molecules that can donate an electron pair to form a covalent bond, but C5H12 (pentane) is a hydrocarbon and does not possess any functional groups that can act as a Lewis base.
The mole ratio of C5H12 to H2 in the reaction is 1:8. This means that for every 1 mole of C5H12, 8 moles of H2 are consumed or produced in the reaction.
The combustion of 2-dimethylpropane (C5H12) in oxygen involves the reaction between the hydrocarbon and oxygen gas to produce carbon dioxide (CO2) and water (H2O) as the main products. The balanced chemical equation for the combustion of 2-dimethylpropane is: C5H12 + 8O2 -> 5CO2 + 6H2O.
To balance the combustion reaction of C5H12 + O2 → CO2 + H2O, first balance the carbon atoms, then the hydrogen atoms, and finally the oxygen atoms. In this case, the balanced equation is: C5H12 + 8 O2 → 5 CO2 + 6 H2O
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Pentane