vì khí N2 tỉ lệ với NO = 1:1
=> nN2 = nNO = giả sử là x mol
PTHH :
10Al + 36HNO3 ----> 10Al(NO3)3 + 18H2O + 3N2
(10/3)x......12x................(10/3)x..............6x.........x
Al + 4HNO3 -----> Al(NO3)3 + 2H2O + NO
..x....4x........................x...........2x..........x
=> \(n_{Al}=\dfrac{10}{3}x+x=0,13\Rightarrow x=0,03\left(mol\right)\)
\(\Rightarrow V=\left(x+x\right)\cdot22,4=\left(0,03+0,03\right)\cdot22,4=1,344\left(l\right)\)
\(13Al+48HNO_3-->13Al\left(NO_3\right)_3+3N_2+3NO+24H_20\)
0.13 0,03 0,03 (Mol)
X3 \( Al^0-->Al^{+3}+3e \)
X13 \(3N^{+5}+13e-->N^{_20}+N^{+2}\)
V= 0,06.22,4 = 1,344 (l)
