Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
+) TH1: R<H
\(n_{H2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
2/15____________________0,2
\(\Rightarrow n_R=\frac{n_{Al}}{2}=\frac{1}{15}\left(mol\right)\)
\(m_{Al}=\frac{2}{15}.27=3,6\left(g\right)\Rightarrow m_R=0,3\left(mol\right)\)
\(\Rightarrow M_R=\frac{0,3}{\frac{1}{15}}=4,5\left(loai\right)\)
+) TH2 : R > H
\(\left\{{}\begin{matrix}n_{Al}:2a\left(mol\right)\\n_R:a\left(mol\right)\end{matrix}\right.\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
2a_______________________3a
\(2R+2nHCl\rightarrow2RCl_n+nH_2\)
a_____________________n/2a
\(\Rightarrow3a+\frac{1}{2}na=0,2\)
* Nếu \(n=1\Rightarrow a=\frac{2}{35}\)
\(\Rightarrow27.2a+Ra=3,9\)
* Nếu \(n=2\Rightarrow a=0,05\)
\(\Rightarrow27.2a+Ra=3,9\)
\(\Rightarrow R=24\left(Mg\right)\)
* Nếu \(n=3\Rightarrow a=\frac{2}{45}\)
\(\Rightarrow27.2a+Ra=3,9\)
\(\Rightarrow R=33,75\left(loai\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Al}=n_{AlCl3}=2a=0,1\left(mol\right)\\n_{Mg}=n_{MgCl2}=a=0,05\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{Muoi}=m_{AlCl3}+m_{MgCl2}=18,1\left(g\right)\)
Câu 1:
CTHH là MSO4
\(\%SO_4=100\%-29,41\%=70,59\%\)
\(\Rightarrow M_{MSO_4}=\dfrac{32+16\times4}{70,59\%}=136\left(g\right)\)
ta có: \(M_M+32+16\times4=136\)
\(\Leftrightarrow M_M+96=136\)
\(\Leftrightarrow M_M=40\)
Vậy M là canxi Ca
Câu 3:
2Na + Cl2 → 2NaCl
\(n_{Na}=\dfrac{1}{23}\left(mol\right)\)
\(n_{Cl_2}=\dfrac{1}{71}\left(mol\right)\)
Theo PT: \(n_{Na}=2n_{Cl_2}\)
theo bài: \(n_{Na}=\dfrac{71}{23}n_{Cl_2}\)
Vì \(\dfrac{71}{23}>2\) ⇒ Na dư
Theo PT: \(n_{NaCl}=2n_{Cl_2}=2\times\dfrac{1}{71}=\dfrac{2}{71}\left(mol\right)\)
\(\Rightarrow m_{NaCl}=\dfrac{2}{71}\times58,5=1,648\left(g\right)\)
Chọn B