Cu | + | 4HNO3 | → | Cu(NO3)2 | + | 2H2O | + |
2NO2 |
n Cu=19,2.64=0,3(mol)
Theo pthh
n NO2=2n Cu=0,6(mol)
V NO2=0,6.22,4=13,44(l)
Lớp 11 á..bài khá dễ..có thể mk làm sai
\(\text{Cu + 4HNO3}\rightarrow\text{Cu(NO3)2 + 2H2O + 2NO2 }\)
Ta có :
\(\text{nCu=0,3 (mol)}\)
\(\rightarrow\text{nNO2=2.0,3-0,6 (mol)}\)
\(\rightarrow\text{VNO2=0,6.22,4=13,44 (l)}\)