Ethane is C2H6.
The answer is 24,23 L water vapors (for a density of 0,804 g/cm3).
88
At 22.4 liters a mole at STP, and the molar mass of Cl2 being 71, 17.32 g is about .242 moles. Multiply the moles by standard volume and you get 5.43 liters.
13000, liters a day but only about 2 and half comes out
1,100 litres is equivalent to 290.589 US gallons (rounded), regardless of what's in the liters. Even if they're completely empty.
60 grams.
Ethane is C2H6.The answer is 24,23 L water vapors (for a density of 0,804 g/cm3).
88
If 15 liters of propane are completely consumed 90,25 grams of carbon dioxide are produced.
2c2h6+702=4co2+6h20 2moles of ethane :7 moles of oxygen 0.9333 of ethane:? cross multiply it and the answer will be 3.266666
at STP 1 mole occupies 22.4 litres. 64.28 / 22.4 is 2.8696428 moles. Multiply this by avagadro's constant (6.022*10^23) gives 1.7281x10^24 molecules
7 liters approximately
Why don't you do your own school assignment? Just take the Carbon part of ethane. We will ignore the Hydrogen that turns to water. When carbon burns, it will produce carbon dioxide. If one molecule contains two carbon atoms, that molecule will burn to produce two molecules of carbon dioxide. Now, you figure it out. To answer this question we need to know what the experimental temperature and pressures are and if ethane is to be considered an ideal gas. STP? RTP? That is not what your question asks. Your question asks, "How many liters of carbon dioxide will be produced. I would assume that it is talking about the same pressure, however I would not know. I would guess that the real issue is asking how much carbon dioxide will be dissolve in the water to form a weak acid and how much will remain free gas. Without more information, Ethane should be considered to be an ideal gas as should water above 100c. You can guess at the temperature by looking up the delta H and calculating with a formula. ---------------------------------------------------------- Lets suppose that the experiment is conducted at STP and that Ethane is an ideal gas. Ethane oxidizes via the formula C2H6 + 7/2 O2 -> 2 CO2 + 3H2O at STP (0'C) water will become a liquid and not contribute significantly to the final volume. At STP pressures are constant and all reactions are exothermic so T remains at 0'C (after everything settles down that is). And lets say that the water remains pure, without significant levels of Carbonic Acid. The question stated that there were 269 litres of Ethane so lets suppose that we are ignoring the initial oxygen gases as well, only considering the volume occupied by the ethane. Oxidized Ethane will produce double the volume of Carbon Dioxide so, at the very simplest, the final volume of Carbon Dioxide with be double the initial volume (of Ethane) so that would be 538 litres. This question gets a lot more complicated if you consider that real behavior of non-ideal gases and that it all takes place in a bomb calorminitor at constant pressure.
2.0 billion gallons
it goes from 40 to 65L
All one to one here. NaOH + HCl --> NaCl + H2O So, getting the moles of either product tell you how many moles H2O produced. Molarity = moles of solute/Liters of solution moles of solute = Liters of solution * Molarity ( 30.00 mL = 0.03 liters ) Moles H2O = 0.03 Liters * 1 M 0.030 moles H2O produced =================
The volume of carbon dioxide is 91,9 L at 0 oC.
At 22.4 liters a mole at STP, and the molar mass of Cl2 being 71, 17.32 g is about .242 moles. Multiply the moles by standard volume and you get 5.43 liters.