Ta có :
\(M_A=\frac{m_{NO}+m_{N_2O}}{n_{NO}+n_{N_2O}}=16,75.2=33,5\)
<=> \(33,5=\frac{30n_{NO}+44n_{N_2O}}{n_{NO}+n_{N_2O}}\)
<=> \(3,5n_{NO}=10,5n_{N_2O}\)
=> \(n_{NO}=3n_{N_2O}\)
Lại có
Al0+ ----> -3e Al3+ N5+----> +3e N2+
0,17_____0,51__0,17 3a______9a____3a
2N5+----> +8e 2N1+
a_______8a____a
=> \(n_{Al}=\frac{4,59}{27}=0,17\left(mol\right)\)
Gọi a là số mol của \(N_2O\)
=> 3a___________NO
=> 9a+8a=0,51
=>a=0,03(mol)
a)=> Σnh2=0,03+0,03.3=0,12(mol)
=> Vh2=0,12.22,4=2,688(l)
b)\(M_B=16,4.2=32,8\left(\frac{gam}{mol}\right)\)
<=> \(\frac{44n_{N_2O}+30n_{NO}}{n_{N_2O}+n_{NO}}=32,8\)
<=>11,2\(n_{N_2O}\)=2,8\(n_{NO}\)
=> \(n_{NO}=4n_{N_2O}\)
mà trước khi trộn : \(n_{NO}=3n_{N_2O}\)
=> trộn thêm số mol khí \(n_{NO}=n_{N_2O}=0,03\left(mol\right)\)
=> \(V_{NO}=0,03.22,4=0,672\left(mol\right)\)
