C3h8 + 5o2 --> 3co2 + 4h2o
C3h8 + 5o2 -> 3co2 + 4h2o
To balance the combustion reaction of propane (C3H8), which produces carbon dioxide (CO2) and water (H2O), we start with the unbalanced equation: C3H8 + O2 → CO2 + H2O. Balancing it gives us: C3H8 + 5 O2 → 3 CO2 + 4 H2O. Therefore, the coefficient for O2 in the balanced equation is 5.
NO. There is no chlorine on the reactant side, so it cannot be balanced.
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.
5
The chemical equation C3H8 + 5O2 -> 3CO2 + 4H2O represents the combustion of propane (C3H8) with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O).
The balanced chemical equation is C3H8 + 5 O2 ⟶ 3 CO2 + 4 H2O. This means that when one molecule of propane (C3H8) reacts with five molecules of oxygen (O2), it produces three molecules of carbon dioxide (CO2) and four molecules of water (H2O).
You mean balance it? 2C2H6 + 7O2 -> 4CO2 + 6H2O
This is the chemical reaction for burning of propane.
A balanced* equation for the burning of propane is: C3H8 + 5 O2 -> 3 CO2 + 4 H2O. *Note that the participial form of "balance" is required for proper grammar in this sentence.
C+co2 = 2co
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.