6,49 moles of water are obtained.
9
First write down the BALANCED reaction eq'n. Octane + oxygen = Water + Carbon Dioxide. 2CH3(CH2)6CH3 + 25O2 = 18H2O + 16CO2 The molar ratios are 2:25 :: 18:16 So '2' moles of octane produces 18 moles of water. So by equivlance 0.468 : x :: 2 : 18 Algebraically rearrgange x = (0.468/2) X 18 => x = 0.234 x 18 = > x = 4.212 moles water produces.
To calculate the mass of octane burned, we can use the heat of combustion of octane which is 5470 kJ/mol. First, convert the given energy to kilojoules per mole. Then, use the molar mass of octane to convert moles to grams. This will give you the mass of octane that must be burned.
To determine the number of moles of carbon dioxide produced when hexane is burned, we need to consider the balanced chemical equation for the combustion of hexane, which is C6H14 + 19/2 O2 -> 6 CO2 + 7 H2O. From the balanced equation, we can see that for every mole of hexane burned, 6 moles of carbon dioxide are produced. Therefore, if 84.4 moles of hexane is burned, 6 * 84.4 = 506.4 moles of carbon dioxide would be produced.
The balanced chemical equation for the combustion of octane is: 2 C8H18 + 25 O2 -> 16 CO2 + 18 H2O From the equation, 1 mole of octane produces 16 moles of CO2. Therefore, 15.0 g of octane will produce 15.0 g x (16/114.20) = 2.10 g of CO2 when burned with 15.0 g of oxygen gas.
9
First write down the BALANCED reaction eq'n. Octane + oxygen = Water + Carbon Dioxide. 2CH3(CH2)6CH3 + 25O2 = 18H2O + 16CO2 The molar ratios are 2:25 :: 18:16 So '2' moles of octane produces 18 moles of water. So by equivlance 0.468 : x :: 2 : 18 Algebraically rearrgange x = (0.468/2) X 18 => x = 0.234 x 18 = > x = 4.212 moles water produces.
To calculate the mass of octane burned, we can use the heat of combustion of octane which is 5470 kJ/mol. First, convert the given energy to kilojoules per mole. Then, use the molar mass of octane to convert moles to grams. This will give you the mass of octane that must be burned.
30 moles
To determine the number of moles of carbon dioxide produced when hexane is burned, we need to consider the balanced chemical equation for the combustion of hexane, which is C6H14 + 19/2 O2 -> 6 CO2 + 7 H2O. From the balanced equation, we can see that for every mole of hexane burned, 6 moles of carbon dioxide are produced. Therefore, if 84.4 moles of hexane is burned, 6 * 84.4 = 506.4 moles of carbon dioxide would be produced.
The balanced chemical equation for the combustion of octane is: 2 C8H18 + 25 O2 -> 16 CO2 + 18 H2O From the equation, 1 mole of octane produces 16 moles of CO2. Therefore, 15.0 g of octane will produce 15.0 g x (16/114.20) = 2.10 g of CO2 when burned with 15.0 g of oxygen gas.
how many moles are in 95.0 gram of octane?
Balanced equation: 2C8H18 + 25O2 ==> 16CO2 + 18H2Omoles of octane used: 325 g x 1 mole/114g = 2.85 moles octanemoles H2O produced: 18 moles H2O/2 moles C8H18 x 2.85 moles C8H18 = 25.65 moles H2O
In 5 moles of octane, C8H18, there are 40 moles of carbon atoms (5 moles octane x 8 carbon atoms) and 90 moles of hydrogen atoms (5 moles octane x 18 hydrogen atoms).
The balanced equation for the combustion of ethane (C2H6) is: 2C2H6 + 7O2 -> 4CO2 + 6H2O From the equation, for every 2 moles of ethane burned, 4 moles of CO2 are produced. Therefore, if 5.60 mol of ethane are burned, then (5.60 mol / 2 mol ethane) * 4 mol CO2 = 11.2 mol of CO2 are produced.
There are 18 moles of water produced in the reaction. This is determined by the stoichiometry of the balanced chemical equation, which shows that for every 2 moles of C8H18 consumed, 18 moles of H2O are produced.
To determine how many moles of octane are present in 16.0 g, you would divide the mass of octane by its molar mass. The molar mass of octane (C8H18) is approximately 114.23 g/mol. Therefore, 16.0 g ÷ 114.23 g/mol = 0.14 moles of octane.