Question

Answer 1

\(n_{NO}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 3Cu + 8HNO₃ --> 3Cu(NO₃)₂ + 2NO + 4H₂O

          0,45<------------------------------0,3

=> m = 0,45.64 = 28,8 (g)

Answer 2

\(n_{NO}=\dfrac{V_{NO}}{22,4}=\dfrac{6,72}{22,4}=0,3mol\)

\(3Cu+8HNO_3\left(loãng\right)\rightarrow3Cu\left(NO_3\right)_2+2NO+4H_2O\)

 0,45                                                          0,3                     ( mol )

\(m_{Cu}=n_{Cu}.M_{Cu}=0,45.64=28,8g\)