Aluminium nitrate is acidic as its made by a weak base with a relatively stronger acid.
Hope that helps ;)
Make sure that you understand what they are asking you from this question, as it can be confusing, but the solution is quite simple. They are stating that they want you to calculate the final concentration of 6.0M HCl once a dilution has been made from 2.0 mL to 500.0 mL. They have given us three values, the initial concentration, initial volume and the final volume. So, we are able to employ the following equation:
C1V1 = C2V2
(6.0M)(2.0mL) = C2(500.0mL)
Therefore, the final concentration, C2 = 0.024M.
40g(1mol NH3/17.04g per mol), N2's molar mass=28.02g
3CuO+2NH3=>3Cu+N2+3H2O
(2.3473mol NH3)(1 mol N2/2 mol NH3)(28.02g/1mol)=32.8873g N2
15.5/32.8873=47.1306% yield