\(M_X=\dfrac{32,8}{0,2}=164\left(\dfrac{g}{mol}\right)\)
Theo đề bài, ta có: \(m_{Ca}:m_N:m_O=10:7:24\)
\(\rightarrow n_{Ca}:n_N:n_O=\dfrac{10}{40}:\dfrac{7}{14}:\dfrac{24}{16}=1:2:6\\ \rightarrow\left(Ca\left(NO_3\right)_2\right)_n=164\\ \rightarrow n=1\)
CTHH: Ca(NO3)2
Đúng 1
Bình luận (0)