Since oxygen is diatomic it requires 2 moles of oxygen.
To find the amount of oxygen required for the combustion of liquid benzene, you would need to balance the combustion reaction equation for benzene. Calculate the moles of benzene from the given mass, then use the stoichiometry of the balanced equation to determine the moles of oxygen required. Finally, convert the moles of oxygen to liters using the ideal gas law.
The ratio of propane to oxygen is 1:5. So for every mole of propane, 5 moles of oxygen gas are required for the complete combustion of propane.Balanced equation:C3H8 + 5O2 --> 3CO2 + 4H2O
3,7 moles of C8H11NO2 have 3,7 moles of oxygen (O2).
2C2H2 + 5O2 ---->4CO2 + 2H2O so 1 mole of acetylene (ethyne) reacts with 2.5 moles of oxygen. Answer 2.5 moles
The balanced chemical equation for the combustion of propane is: C3H8 + 5 O2 -> 3 CO2 + 4 H2O. This means that 5 moles of oxygen are required to completely combust 1 mole of propane. Therefore, 20 moles of oxygen would be produced from the complete combustion of 4 moles of propane.
0. Hydrogen doesn't "reackt" to form Nitrogen Monoxide.
800 g oxygen are needed.
999 g
16 grams of oxygen how many moles is 0,5 moles.
To find the amount of oxygen required for the combustion of liquid benzene, you would need to balance the combustion reaction equation for benzene. Calculate the moles of benzene from the given mass, then use the stoichiometry of the balanced equation to determine the moles of oxygen required. Finally, convert the moles of oxygen to liters using the ideal gas law.
The ratio of propane to oxygen is 1:5. So for every mole of propane, 5 moles of oxygen gas are required for the complete combustion of propane.Balanced equation:C3H8 + 5O2 --> 3CO2 + 4H2O
The balanced equation for the reaction is 2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O. This shows that 13 moles of diatomic oxygen are required to burn 2 moles of butane. By proportionality, (4.8/2)13 or 31.2 moles of oxygen are required to burn 4.8 moles of butane. This corresponds to 31.2(32) or 1.0 X 103 grams of oxygen.
C3H8 + 5O2 --> 3CO2 + 4H2O 2.75 mole C3H8 (5 moles O2/1 mole C3H8)(32 grams/1 moleO2) = 440 grams oxygen required =====================
3,7 moles of C8H11NO2 have 3,7 moles of oxygen (O2).
The nunber of moles of oxygen is 2,5.
Assuming complete combustion of butane, you need 15 moles of oxygen to react with 5 moles of butane according to the balanced chemical equation: [ 2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O ]
2C2H2 + 5O2 ---->4CO2 + 2H2O so 1 mole of acetylene (ethyne) reacts with 2.5 moles of oxygen. Answer 2.5 moles