If 4.00 kg of FeO3 are avaiable to react, how many moles of Co are needed? How man moles of each product are formed?
any dumbass can learn how to ride a bike, its the smart ones that know how to react and ride safely. *edit* and i agree somewhat about MSF. MSF courses are really only beneficial for people that have little to no experience on a bike what so ever.
as with driving a car and a bike. over confidence is a killer and just because someone have never had an accident only show how well the other drivers are to miss him. he need to work it out fast before it costs him his life
The first one is clearly not “like this” –?he has crashed three times that you know of. However, you should be wary of judging other riders, there is another rider stereotype – all the gear, no idea. Courses and gear are useful and important, they do not, cannot, give you everything you need to know. If you are using them to teach you, then you should probably look elsewhere.
Fe2O3 + 3 CO → 2 Fe + 3 CO2 (4000 g Fe2O3) / (159.6882 g Fe2O3/mol) x (3 mol CO / 1 mol Fe2O3) 75.1 mol CO
1)Balance the reaction Fe2O3 3CO --- 2Fe 3 CO2 2) Convert 4 kg into grams you get .004 grams Fe2O3 3)Divide by 159 g/mol (the Molecular weight of Iron(III) oxide) you will get a very small number 2.51 x 10^-5 mol 4) For every mole of Iron (III) oxide you need 3 moles of Carbon Monoxide so just multiply by 3 and you will get 7.5 x 10^-5 moles
