\(n_{Zn}=\dfrac{6,5}{65}=0,1mol\)
\(n_{H_2SO_4}=\dfrac{100}{98}=1,02mol\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
0,1 < 1,02 ( mol )
0,1 0,1 0,1 ( mol )
\(m_{ZnSO_4}=0,1.161=16,1g\)
\(V_{H_2}=0,1.22,4=2,24l\)
\(m_{ddspứ}=6,5+100-0,1.2=106,3g\)
\(C\%_{ZnSO_4}=\dfrac{16,1}{106,3}.100=15,14\%\)
Đúng 2
Bình luận (1)