Question

Answer 1

\(n_{NO_2}=\dfrac{20,16}{22,4}=0,9\left(mol\right)\)

PTHH: Al + 6HNO₃ --> Al(NO₃)₃ + 3NO₂ + 3H₂O

____a------------------------->a---->3a

Cu + 4HNO₃ --> Cu(NO₃)₂ + 2NO₂ + 2H₂O

b---------------------->b---------->2b

=> \(\left\{{}\begin{matrix}27a+64b=15\\3a+2b=0,9\end{matrix}\right.=>\left\{{}\begin{matrix}a=0,2\\b=0,15\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%Al=\dfrac{27.0,2}{15}.100\%=36\%\\\%Cu=\dfrac{0,15.64}{15}.100\%=64\%\end{matrix}\right.\)

b) 

PTHH: 2Al(NO₃)₃ + 4Ba(OH)₂ --> 3Ba(NO₃)₂ + Ba(AlO₂)₂ + 4H₂O

_______0,2--------->0,4

Cu(NO₃)₂ + Ba(OH)₂ --> Ba(NO₃)₂ + Cu(OH)₂\(\downarrow\)

_0,15------->0,15

=> n_{Ba(OH)2 (pư)} = 0,4 + 0,15 = 0,55 (mol)

=> n_{Ba(OH)2 (thực tế)} = \(\dfrac{0,55.120}{100}=0,66\left(mol\right)\)

=> \(V_{dd}=\dfrac{0,66}{1}=0,66\left(l\right)\)