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I started my problem by writing a balanced equation of the reaction but now I haven't the slightest idea of what to do with it. 2H2+O2->2H20
How many moles are in 39.7 g of Hydrogen? How many moles of hydrogen are in 355 g of water? How many moles of water are in 355 g of water? We need these three values to determine if the combustion process is stoichiometric.
How would I figure the moles H/355g H20?
isn't simply a matter of 355 - 39.7 g oxygen
355 g of water = 355/18 moles
Did you figure out the number of moles of hydrogen and water?
Hydrogen+Oxygen->Water If you burn 39.7 g of hydrogen and produce 355g of water, how much oxygen reacted? so you have 355 grams of water in the end or the process Find the g/mol of H2O 355g H2O *(Oxygen g/mol/H2O g/mol) = grams of oxygen you reacted
To find the g/mol of H2O simply (hydrogen g/mol)*2 + (g/mol of oxygen) = H2O g/mol
355g H2O *(15.9994 g/mol/18.0153 g/mol) = 315.28 grams
of Oxygen were reacted theoretically of course assuming all the oxygen was pushed forward to water.
if you look at the formula you will notice the g/mol cross out and the H2O and the H2O cross out so all you are left with is Oxygen and grams
Hope this helps
355 - 39.7 = 315.3 g
jimmy rep your answer is just as good as mine but what i have provided her is much more universal for dealing with problems like this that are a little more complex but yeah.
It's helping but I'm getting confused. I know that there's 39.7mol H per 39.7g H You said to do 39.7g*2=79.4 79.4+(g/mol of oxygen) = H2O g/mol Now I'm just guessing 79.4+16=95.4 Sorry guys it's been a few years since I've done chemistry
yea - i agree
from equation and using approximate moles of reactants 2 H2 = 4 (4*1) 02 = 32 (2*16) 2 H20 = 36 (4+32) ratio 1:8:9 39.7 g of hydrogen reacts with 8*39.7 g oxygen giving 9* 39.7 g water
Okay, now I get it. At first I solved it the same as Jimmy because of the Law of Conservation of Mass but I figured I should learn the proper way. Thank you to all who helped!!