21.10 grams of magnesium react with excess sulfuric acid. How many liters of hydrogen are produced?

1 Answer
anor277
Feb 22, 2016

We have (almost!) a mole of magnesium atoms. So we generate, almost a mole of dihydrogen gas, approx #25# #dm^3#!

Explanation:

#Mg(s) + H_2SO_4(aq) rarr MgSO_4(aq) + H_2(g)uarr#

Given the equation (which you should be able to reproduce) there is a 1:1 equivalence of magnesium metal to dihydrogen gas.

Moles of #Mg# #=# #(21.10*g)/(26.98*g*mol^-1)# #=# #??# #mol#.

Now it is a fact, that at #1# atmosphere pressure and #298# #K#, #1# #mol# of ideal gas (and we assume dihydrogen approximates ideality!) occupies #24.5# #dm^3# (#1# #dm^3# #=# #1# #L#).

So volume of dihyfrogen (at #SLC#) #=#

#(21.10*cancel(g))/(26.98*cancel(g)*cancel(mol^-1))# #xx# #24.5*L*cancel(mol^-1)# #=# #??L#

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