Assuming that the water is produced by complete combustion of propane, the balanced equation is:
C3H8 + 5 O2 = 3 CO2 + 4 H2O.
The gram molecular mass of propane is three times the gram Atomic Mass of carbon plus eight times the gram atomic mass of hydrogen = 3(12.011) + 8(1.008) = 44.097, and the gram molecular mass of water is the gram atomic mass of oxygen plus twice the gram atomic mass of hydrogen = 15.999 + 2(1.008) = 18.015. The balanced reaction equation shows that four molecules of water are produced for each molecule of propane, so that the ratio of grams of propane reacted to grams of water produced is 44.097/(4 X 18.015) = 0.61191, to the justified number of significant digits. Therefore, the amount of propane required to produce 8.00 grams of water = 8.00(0.61191) = 4.90 grams to the clearly justified number of significant digits, matching the three significant digits given for 8.00, or 4.895 grams, where the last digit is depressed because it may well not be accurate within + 1
To determine the moles of water produced from the reaction of 6.00 grams of propane, first calculate the moles of propane using its molar mass. Then, use the balanced chemical equation to find the moles of water produced based on the stoichiometry of the reaction.
When 2.5 moles of oxygen react with hydrogen, they react in a 1:2 ratio to produce water. Therefore, 2.5 moles of oxygen will produce 5 moles of water. To convert moles to grams, you'll need to know the molar mass of water, which is approximately 18 grams/mol. So, 2.5 moles of oxygen will produce 90 grams (5 moles x 18 grams/mole) of water.
Calculate the mass in grams of water vapor produced if 3.11 moles of propane is burned
Let's see. C3H8 + 5O2 -> 3CO2 + 4H2O For every one molecule of propane burned there is four molecules of water produced. Or, this is the actuality. 1 molecule propane (1 mole C3H8/6.022 X 1023)(4 mole H2O/ 1mole C3O8) = 6.64 X 10 -24 molecules water
No, two acids cannot react to produce a base. Acids react with bases to produce salt and water through a neutralization reaction.
To determine the moles of water produced from the reaction of 6.00 grams of propane, first calculate the moles of propane using its molar mass. Then, use the balanced chemical equation to find the moles of water produced based on the stoichiometry of the reaction.
When 2.5 moles of oxygen react with hydrogen, they react in a 1:2 ratio to produce water. Therefore, 2.5 moles of oxygen will produce 5 moles of water. To convert moles to grams, you'll need to know the molar mass of water, which is approximately 18 grams/mol. So, 2.5 moles of oxygen will produce 90 grams (5 moles x 18 grams/mole) of water.
Calculate the mass in grams of water vapor produced if 3.11 moles of propane is burned
7
Let's see. C3H8 + 5O2 -> 3CO2 + 4H2O For every one molecule of propane burned there is four molecules of water produced. Or, this is the actuality. 1 molecule propane (1 mole C3H8/6.022 X 1023)(4 mole H2O/ 1mole C3O8) = 6.64 X 10 -24 molecules water
155.2 g
No, two acids cannot react to produce a base. Acids react with bases to produce salt and water through a neutralization reaction.
To find the grams of nitrogen dioxide needed, first calculate the moles of nitrogen monoxide using Avogadro's number. Then, use the balanced chemical equation to determine the moles of nitrogen dioxide required. Finally, convert moles to grams using the molar mass of nitrogen dioxide.
The end products of burning propane (C3H8) with 100-percent efficiency are carbon dioxide (CO2) and water (H2O). The complete combustion of propane occurs when the propane molecules react with oxygen to produce these two compounds along with heat energy.
yes it does.
For the reaction 2H₂ + O₂ → 2H₂O, we know that the molar ratio of H₂ to O₂ is 2:1. To produce 900 grams of water, we need 450 grams of hydrogen (900g / 2). Therefore, we need to add 450 grams of hydrogen to 800 grams of oxygen to produce 900 grams of water.
C3H8 + 5O2 -> 3CO2 + 4H2O That is the complete combustion for Propane.