What volume of 0.250 M HNO3(aq) is required to completely react with 10.0 g of zinc metal according to the equation: 2 HNO3(aq) + Zn(s) → Zn(NO3)2(aq) + H2(g) ?

What volume of 0.250 M HNO3(aq) is required to completely react
with 10.0 g of zinc metal according to the equation:
2 HNO3(aq) + Zn(s) → Zn(NO3)2(aq) + H2(g) ?

1 Answer
May 23, 2016

Answer:

#Zn(s) + 2HNO_3(aq) rarr Zn(NO_3)_2(aq) + H_2(g)uarr#

We need over #1*L# of nitric acid.

Explanation:

The stoichiometric equation shows a #2:1# equivalence between the acid and zinc respectively.

#"Moles of Zinc"# #=# #(10.0*g)/(65.38*g*mol^-1)# #=# #0.153*mol#.

Thus we need a #0.306# molar quantity of acid.

We have #0.250*mol*L^-1# nitric acid available, thus,

#(0.306*cancel(mol))/(0.250*cancel(mol)*cancel(L^-1))xx1000*mL*cancel(L^-1)# #=# #??*mL#