Câu 1:
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
PTHH: Al + 6HNO3 (đặc, nóng) ---> Al(NO3)3 + 6NO2 + 3H2O
0,1 0,6
\(\rightarrow V_{NO_2\left(đktc\right)}=0,6.22,4=13,44\left(l\right)\)
Câu 2:
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: Fe + 2HCl ---> FeCl2 + H2
0,1 0,1
\(\rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{5,6}{8,8}=63,63\%\\\%m_{Cu}=100\%-63,63\%=36,37\%\end{matrix}\right.\)
Đúng 2
Bình luận (0)
