When #10*g# dihydrogen gas, and #32*g# dioxygen gas react, which is the limiting reagent. And what mass of water will result?

1 Answer
Jan 26, 2017

Answer:

You need (i) a stoichiometrically balanced equation...........

Explanation:

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

And (ii) equivalent quantities of dioxygen and dihydrogen.

#"Moles of dihydrogen"# #=# #(10*g)/(2.02*g*mol^-1)~=5*mol#.

#"Moles of dioxygen"# #=# #(32*g)/(32.0*g*mol^-1)~=1*mol#

Now, clearly, #O_2# is the limiting reagent, and there is a stoichiometric excess of #H_2#. At most #2 *mol# of #H_2# react quantitatively with the #1*mol# of #O_2#, to give #2*mol# of #H_2O#, which represents a mass of #36*g#.

At given temperature and pressure what volume of #"dioxygen"# will be required for a given volume of #"dihydrogen"#?