Question

The equation for the reaction of 6.5 g of zinc granules with 0.28 moles of nitric acid is as follows: 3Zn + 8HNO3  3Zn(NO3)2 + 2NO + 4H2O can some one do it please?

(i)Calculate the number of moles of zinc (Zn) used in the reaction.
(ii) Identify the reactant that is limiting in the reaction.
(iii) Calculate the volume of nitrogen monoxide in cm3
, at room temperature and pressure,
produced in the reaction.
(iv) What mass of zinc nitrate (Zn(NO3)2) was produced in the reaction?
(v) How many molecules of water were produced in the reaction?

Answer 1

I'll answer each part, in turn. This is a very thorough problem that demands a mastery of stoichiometry, it isn't challenging, but is great practice.

Explanation:

Consider,

`3Zn(s) + 8HNO_3(aq) to 3Zn(NO_3)_2(aq) + 2NO(g) + 4H_2O(l)`

In this case,

`6.5g * ("mol")/(65.4g) approx 9.9*10^-2"mol"`

of zinc are reacted, and,

`9.9*10^-2"mol" * (8HNO_3)/(3Zn) approx 0.265"mol"`

are required to completely react with this amount of zinc.

Recall, we have a molar amount greater than this of nitric acid, so zinc is the limiting reactant per deduction.

Moreover,

`9.9*10^-2"mol" * (2NO)/(3Zn) approx 6.6*10^-2"mol"`

`1.0atm * V = 6.6*10^-2"mol" * (0.08206L * atm)/("mol" * K) * 298K`
`therefore V approx 1.62L * (10^3mL)/L * (cm^3)/(mL) = 1.61*10^3cm^3`

of nitrogen monoxide gas,

`9.9*10^-2"mol" * (3Zn(NO_3)_2)/(3Zn) approx 9.9*10^-2"mol" * (189.4g)/("mol") approx 19g`

equivalent of aqueous nitrate, and,

`9.9*10^-2"mol" * (4H_2O)/(3Zn) * (6.02*10^23"molecules")/"mol" approx 7.9*10^24`

molecules of water are produced.