4 C3H8 + 20 O2 ---> 12 CO2 + 16 H2O
So 20 mol of O2 are needed.
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 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.
For every 1 mole of propane burned, 5 moles of oxygen are required. This means that 44 grams of propane requires 160 grams of oxygen to burn completely. Therefore, 100 grams of propane would require (100 grams propane * 160 grams oxygen / 44 grams propane) = 363.64 grams of oxygen to burn completely.
To find the number of moles of propane in the container, first convert the mass of propane to grams: 2.55 kg = 2550 grams Next, calculate the number of moles using the molar mass of propane: Molar mass of C3H8 = 44.11 g/mol Number of moles = 2550 g / 44.11 g/mol = 57.86 moles Therefore, there are 57.86 moles of propane in the pressure container.
The balanced chemical equation for the reaction between hydrogen and oxygen is: 2H2 + O2 -> 2H2O This equation shows that 1 mole of O2 reacts with 2 moles of H2. So, to completely react with 6 moles of H2, you would need 3 moles of O2.
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 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.
For every 1 mole of propane burned, 5 moles of oxygen are required. This means that 44 grams of propane requires 160 grams of oxygen to burn completely. Therefore, 100 grams of propane would require (100 grams propane * 160 grams oxygen / 44 grams propane) = 363.64 grams of oxygen to burn completely.
The balanced chemical equation for the combustion of 2 moles of propane (C3H8) is 2C3H8 + 7O2 → 6CO2 + 8H2O. This equation shows that 16 moles of oxygen gas (O2) are needed to react completely with 2 moles of propane.
Propane is C3H8 and the combustion equation is C3H8 + 5O2 ==> 3CO2 + 4H2OSo the complete combustion of 1 mole of propane requires 5 moles of oxygen.
The answer is: 5 moles oxygen and 4 moles H2O.The reaction is:C3H8 + 5 O2 = 3 CO2 + 4 H2O
30 moles
The combustion of propane (C3H8) can be represented by the balanced chemical equation: (C3H8 + 5O2 \rightarrow 3CO2 + 4H2O). According to this equation, 1 mole of propane requires 5 moles of oxygen for complete combustion. Therefore, 5 moles of oxygen are needed for the reaction when burning 1 mole of propane.
0. Hydrogen doesn't "reackt" to form Nitrogen Monoxide.
Since propane has the formula C3H8, each mole of propane will have 8 moles of hydrogen atoms, so 5 moles of propane will contain 5x8=40 moles of hydrogen.
To determine how many moles of oxygen are needed to completely react with isooctane (C₈H₁₈), we can use the balanced combustion reaction: 2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O. From this equation, 2 moles of isooctane react with 25 moles of oxygen, indicating that 1 mole of isooctane requires 12.5 moles of oxygen. Therefore, for 4 moles of isooctane, 4 × 12.5 = 50 moles of oxygen are needed.
5.8