Question

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) ?

Answer 1

`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`