When 1.57 mol #O_2# reacts with #H_2# to form #H_2O#, how many moles of #H_2# are consumed in the process?

1 Answer
Dec 9, 2016

Answer:

Look at the stoichiometry, the chemical equivalence, of the reaction:

#H_2(g) + 1/2O_2(g) rarr H_2O(l)#

Explanation:

#H_2(g) + 1/2O_2(g) rarr H_2O(l)#, or I could simply double the equation,

#2H_2(g) + O_2(g) rarr 2H_2O(l)#

Now, clearly, for each mole/equiv of dioxygen, 2 moles/equivs dihydrogen react. If there are #1.57*mol# #O_2#, stoichiometric equivalence requires #2xx1.57*mol# dihydrogen. (i) How many moles of water will be generated; and (ii) what is the mass of the water?