Question

Hòa tan 4,32g Al trong dd HNO3 2M thu 2,464l hh NO, N2O. Tính V dd HNO3 đã dùng

Answer 1

8Al+ 30HNO3---->8Al(NO3)3+3N2O+15H2O

x---------------------------------------0,375x

Al + 4HNO3----->Al(NO3)3+NO+2H2O

y-------------------------------------y

n\(_{Khí}=\)\(\frac{2,464}{22,4}=0,11\left(mol\right)\)

Theo bài ta có hệ pt

\(\left\{{}\begin{matrix}27x+27x=4,34\\0,375x+y=0,11\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,08\\y=0,08\end{matrix}\right.\)

Theo pthh1

n\(_{HNO3}=\frac{30}{8}n_{Al}=0,3\left(mol\right)\)

Theo pthh2

n\(_{HNO3}=4n_{Al}=0,32\left(mol\right)\)

Σn\(_{HNO3}=0,3+0,32=0,62\left(mol\right)\)

V\(_{HNO3}=\frac{0,62}{2}=0,31l\)

Answer 2

https://i.imgur.com/9Tx3QNi.jpg

Answer 3

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