Balanced equation: 2C8H18 + 25O2 ==> 16CO2 + 18H2O
moles of octane used: 325 g x 1 mole/114g = 2.85 moles octane
moles H2O produced: 18 moles H2O/2 moles C8H18 x 2.85 moles C8H18 = 25.65 moles H2O
Burning octane is an exothermic reaction because it releases energy in the form of heat and light as it reacts with oxygen to form carbon dioxide and water.
CO, CO2, H2O and more comlicated structures.
The formula for normal octane is C8H18. Its molar mass is 114.23 g mol−1 The formula for its combustion is 2C8H18 + 25O2 --> 16CO2 + 18H2O So 1 mole of octane gives 9 moles of water. One mole of water has a mass of 18 g 19.8 g of octane is 114.23/19.8 moles so its combustions gives ((114.23/19.8) x 9 x 18 ) = 934.61 g of water
The oxygen necessary for burning is not sufficient.
OxygenFood (glucose)Starch (excess food/glucose)
octane + oxygen --> water + carbon dioxide
Burning octane is an exothermic reaction because it releases energy in the form of heat and light as it reacts with oxygen to form carbon dioxide and water.
A blue flame would indicate ethane burning in excess oxygen. Blue flames are typically associated with complete combustion and sufficient oxygen supply during the burning process.
The mass of heavy water produced when 7,00grams of oxygen reacts with excess D2 is 7,875 g.
CO, CO2, H2O and more comlicated structures.
Octane + oxygen ---> carbon dioxide + water Lulu
Carbon is the element that is black and is produced when there is incomplete burning.
Carbon Dioxide & Water, Complete burning with excess of oxygen gives out Carbon Dioxide(CO2) + Water , While incomplete burning with limited amount of oxygen gives out Carbon monoxide (CO) + Water
The formula for normal octane is C8H18. Its molar mass is 114.23 g mol−1 The formula for its combustion is 2C8H18 + 25O2 --> 16CO2 + 18H2O So 1 mole of octane gives 9 moles of water. One mole of water has a mass of 18 g 19.8 g of octane is 114.23/19.8 moles so its combustions gives ((114.23/19.8) x 9 x 18 ) = 934.61 g of water
The equation for the burning of sulfur in the presence of oxygen to produce sulfur dioxide is: S (sulfur) + O2 (oxygen) --> SO2 (sulfur dioxide).
The oxygen necessary for burning is not sufficient.
To determine the limiting reactant, compare the moles of each reactant to the stoichiometry of the balanced. In this case, the balanced equation is: 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O The moles ratio between octane (C8H18) and oxygen (O2) is 2:25. Calculate the ratio for each reactant: Octane: 0.400 mol * (25 mol O2 / 2 mol C8H18) = 5.00 mol O2 needed Oxygen: 0.800 mol O2. Since the actual moles of oxygen available (0.800 mol) are greater than the moles needed for the reaction with octane (5.00 mol), oxygen is in excess and octane is the limiting reactant.