If you have 5.0 L of 0.10 M HNO3, then how many grams of Aluminum nitrate can you make?
Balanced Equation: 6 HNO3 + 2 AL ---> 2 AL(NO3)3 + 3 H2
- GodLv 71 month agoFavorite Answer
5L x 0.1M = 0.5 moles of HNO3
6 moles of HNO3 yields 2 moles of Al(NO3)3
0.5 moles of HNO3 yields 0.5 x 2/6 = 0.167 moles of Al(NO3)3
0.167 moles x 213g/mole = 29.2 g of Al(NO3)3
- pisgahchemistLv 71 month ago
Aluminum and nitric acid .....
Whoa, Nellie. Back the boat up. Do you have any evidence that there is a reaction? Or did your teacher simply give you this reaction in a question? If that's the case then your teacher has made a mistake. There is no reaction! You won't make any Al(NO3)3 via this reaction.
Why is there no reaction? You would think (apparently erroneously) that since Al is above H in the activity series that H2 should be a product. There are two reasons why that won't happen.
First, aluminum metal has an passivating layer of Al2O3 on the surface that forms immediately when aluminum is exposed to O2 in the air. The Al2O3 layer prevents further reaction unless the acid is quite concentrated, or unless there is some other reactant which breaks down the passivating layer. Simply adding a source of chloride ion will do that.
Al2O3(s) + 6H+ + 8Cl- --> 2[AlCl4]^- + 3H2O(l)
Once the passivating layer is removed, then the aluminum will be oxidized by the nitrate ion, which brings us to the second reason why H2 is not a product.
Al(s) + 4H+ + NO3^- --> Al3+ + NO(g) + 2H2Ol)
A secondary reaction further oxidizes NO to NO2 when NO reacts with O2 in the air.
2NO(g) + O2(g) --> 2NO2(g)
It's also possible to observe the further reduction of NO3^- to form not only NO2 and NO, but N2 and NH3. The conditions play an important role in the degree of reduction.
Of course, it is possible to do the math for the reaction as written in your original question, but please understand that that reaction will not occur.
2Al(s) + 6HNO3(aq) --> 2Al(NO3)3 + 3H2 ....... (sic)
........ ...... 5.0L .....................?g
........ ...... 0.10M
5.0L x (0.10 mol HNO3 / 1L) x (1 mol Al(NO3)3 / 3 mol HNO3) x (213.0g Al(NO3)3 / 1 mol Al(NO3)3) = 35.5g .....(hypothetically) ...... ideally, the answer is rounded to two significant digits.