Three different elements: C, H, O.
3 C-atoms
8 H-atoms
10 (=3*2+4) O-atoms
The equation you provided is not balanced. To balance it, you need to make sure that the number of each type of atom is the same on both sides of the equation. Once the equation is balanced, you can count the number of reactions by looking at the coefficients of the reactants and products in the balanced equation.
For the reaction of propane (C3H8) with oxygen (O2), the balanced equation is: C3H8 + 5O2 -> 3CO2 + 4H2O. This means that 5 moles of O2 are required to react completely with 1 mole of propane (C3H8). Therefore, to react completely with 4 moles of propane, you would need 20 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
carbon dioxide and water oxygen gas and carbon atoms
This question sound like one that can easily be answered if one has a general chemistry textbook. My advice is to look at doing some unit conversions to find out how many moles of propane are in 36.1 g of propane first. With a balanced chemical equation for combustion, and assuming the propane is the limiting reagent, use dimensional analysis to convert your moles of propane to moles of carbon dioxide.
C-carbon H-hydrogen O-oxygen Three different elements.
The equation you provided is not balanced. To balance it, you need to make sure that the number of each type of atom is the same on both sides of the equation. Once the equation is balanced, you can count the number of reactions by looking at the coefficients of the reactants and products in the balanced equation.
C3H8 + 5O2 --> 3CO2 + 4H2O 2.75 mole C3H8 (5 moles O2/1 mole C3H8)(32 grams/1 moleO2) = 440 grams oxygen required =====================
You have to burn C3H8 in O2. You get 3CO2 plus 4H2O. So to burn one mole of C3H8, you need 5 moles of O2. That means you need one fifth of C3H8 as compared to O2. So you need 0.567/5 = 0.1134 moles of C3H8. Hence the answer.
Propane is burned to provide the heat in many cooking grills. The chemical reaction for this process is shown in the equation below. C3H8 + 5O2 ? 3 CO2 + 4H2O + energy What are the products in this chemical reaction? 3CO2 + 4H2O + energy
Three different elements: C, H, O. 3 C-atoms 8 H-atoms 10 (=3*2+4) O-atoms
For the reaction of propane (C3H8) with oxygen (O2), the balanced equation is: C3H8 + 5O2 -> 3CO2 + 4H2O. This means that 5 moles of O2 are required to react completely with 1 mole of propane (C3H8). Therefore, to react completely with 4 moles of propane, you would need 20 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 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.
For every mole of propane (C3H8) that reacts, 4 moles of water (H2O) are produced. Since 1 mole of gas occupies about 22.4 L at STP, if 4.50 L of propane reacts, you would need to convert the volume of propane to moles using the ideal gas law to determine the moles of water produced.
carbon dioxide and water oxygen gas and carbon atoms
The balanced equation is C3H8 + 5O2 ---> 3CO2 + 4H2O moles C3H8 = 23.7 g x 1 mol/44 g = 0.539 moles moles O2 needed = 5 x 0.539 moles = 2.695 moles O2 (it takes 5 moles O2 per mole C3H8) grams O2 needed = 2.695 moles x 32 g/mole = 86.2 grams O2 needed (3 sig figs)