How do I calculate the amount of water produced by the combustion of 16 g of methane?
To solve such questions you must write a balanced reaction first.CH4 (g)+2O2 (g)————-> CO2 (g) +2H2O (g)Now according to the reaction 1 mole of CH4 gives 2 moles of H2OAnd 16 grams of Methane mean 1 mole (moles=given mass/molecular mass. Molecular mass of methane is 16)so 1 mole of methane = 16/16 (given mass 16 and molecular mass 16)since 1 mole of methane is getting used up to form 2 mole of water.number of moles=given mass/molecular mass ——- eq. 1molecular mass of water is 18.2=x/18 ( where x is the mass of water produced , using equation 1)2*18=36gramsso 36 grams of water will be produced.
Burning ethylene....if 14 grams of ethylene is reacted and the yield of water is 7.84 grams, the percent yield is what?
using the reaction burning ethylene.... C2H4 + 3O2 -----> 2CO2 + 2H2O a.) if 14 grams of ethylene is reacted and the yield of water is 7.84 grams, the percent yield is what? b.) how many moles of carbon dioxide would be formed if you burned 18 moles of ethylene? c.) how many grams of oxygen are required for the complete reaction of 45 grams of ethylene? Thanks so much!
How many moles of O2 are needed to react with 24 moles of C2H6?
Sounds like your typical stoichiometry question! (either that or it’s a typical chemical balancing problem. I can never see the difference).Well, let’s take a step back and see the general problem before calculating the number of mols of O2 you’ll need. You’re exposing ethane molecules to oxygen molecules, and you’re told that these two things will supposedly react. What kind of products can you assume?Well, since a reaction is supposed to rearrange molecules to form different molecules, how about the formation of CO2 and H2O? Your initial compound does seem to have carbon atoms and hydrogen atoms, so let’s start with an equation!C2H6(l) + O2 (g) → CO2 (g) + H2O (l)Now there’s a problem here. Shouldn’t we have the same atoms of everything on both sides? Well then, why do I have two carbon atoms on the left, but one carbon on the left? Why do I have 6 hydrogens on the left, but only two on the right? Why do I have three oxygens on the right, but only two on the left?Let’s balance the equation by saying that the reaction produces TWO CO2 molecules instead of one.C2H6(l) + O2 (g) → 2CO2 (g) + H2O (l)Now we’ve got two carbon atoms on both sides, but now we have even MORE oxygen atoms on top of the imbalance of hydrogen atoms! Well, let’s say that we produce THREE water molecules instead:C2H6(l) + O2 (g) → 2CO2 (g) + 3H2O (l)We balanced out the hydrogens, but now the right side has a whoppin’ total of 7 oxygen atoms. Now what? Well, let’s say we:C2H6(l) + [math]7/2[/math]O2 (g) → 2CO2 (g) + 3H2O (l)And now we have equal number of atoms on both sides! 2 C atoms, 7 O atoms, and 6 H atoms! And if you don’t like fractions, why not multiply the entire thing by two?2C2H6(l) + 7O2 (g) → 4CO2 (g) + 6H2O (l)You got the same number of atoms on both sides, and you don’t have to deal with fractions anymore!Now to get back to your original question. How many moles of O2 do you need for 24 moles of ethane? Well, for every two moles of ethane, it seems like you need 7 mols of O2. If you had 24 mols of ethane, well, that’s 12x more moles than what you have in this reaction, right? Therefore, you’d need 12x more mols of O2 for the same reaction to go to completion, or 84 mols of O2.And there you have it, mate!