the answers is 6 _apex
The balanced equation is 2KHCO3(s) → K2CO3(s) + H2O(g) + CO2(g). The coefficient of carbon dioxide (CO2) in the balanced equation is 1.
To balance the equation C6H14 + O2 -> CO2 + H2O, start by balancing the carbon atoms. This requires putting a coefficient of 6 in front of CO2. Next, balance the hydrogen atoms by adding a coefficient of 7 in front of H2O. Finally, balance the oxygen atoms by adjusting the coefficient in front of O2, which in this case is 9.
The balanced reaction of propane (C3H8) with oxygen (O2) to form carbon dioxide (CO2) and water (H2O) is: C3H8 + 5O2 → 3CO2 + 4H2O
Burning of propane:CH3H8 + 5 O2 = 3 CO2 + H2OBurning of butane:2 CH4H10 + 13 O2 = 8 CO2 + 10 H2O
5
C3H8 + 5O2 = 3CO2 + 4H2O, so the coefficient for O2 is 5
the answers is 6 _apex
2c4h10 + 13o2 => 8co2 + 10h2o (I am having some trouble with my typography today, but all those letters above should be capitalized.)
To calculate the amount of CO2 produced when burning 34.3 grams of C3H8 (propane), you need to balance the chemical equation for the combustion of C3H8. Since each mole of C3H8 produces 3 moles of CO2, you first convert 34.3 grams of C3H8 to moles, calculate the moles of CO2 produced, and then convert that to grams of CO2.
If 100.0g of C3H8 is burned completely, all the carbon in C3H8 will form CO2. The molar mass of C3H8 (propane) is 44.1 g/mol, and it contains 3 carbons. This means 100.0g of C3H8 contains 3*(12.01g) carbon atoms, which will produce 3*(44.01g) of CO2, resulting in 132.03g of CO2 being produced.
The balanced equation is 2KHCO3(s) → K2CO3(s) + H2O(g) + CO2(g). The coefficient of carbon dioxide (CO2) in the balanced equation is 1.
To calculate the grams of CO2 produced by burning 22 grams of C3H8, first determine the moles of C3H8 using its molar mass, then use the balanced chemical equation for the combustion of C3H8 to find the moles of CO2 produced, and finally convert moles of CO2 to grams using the molar mass of CO2.
Complete combustion of propane (C3H8) produces 3 moles of CO2 for every 1 mole of C3H8 burned. Therefore, when 2.13 moles of C3H8 burn, 6.39 moles of CO2 will be formed.
To balance the equation C6H14 + O2 -> CO2 + H2O, start by balancing the carbon atoms. This requires putting a coefficient of 6 in front of CO2. Next, balance the hydrogen atoms by adding a coefficient of 7 in front of H2O. Finally, balance the oxygen atoms by adjusting the coefficient in front of O2, which in this case is 9.
C3h8 + 5o2 -> 3co2 + 4h2o
The balanced reaction of propane (C3H8) with oxygen (O2) to form carbon dioxide (CO2) and water (H2O) is: C3H8 + 5O2 → 3CO2 + 4H2O