\(2Al + 6HCl \rightarrow 2AlCl_3 + 3H_2\)

\(n_{Al}= \dfrac{5,4}{27}= 0,2 mol\)

Theo PTHH:

 \(n_{AlCl_3}= n_{Al}= 0,2 mol\)

\(\Rightarrow m_{AlCl_3}= 0,2 . 133,5=26,7 g\)

Theo PTHH: 

\(n_{H_2}= \dfrac{3}{2} n_{Al}= 0,3 mol\)

\(\Rightarrow V= 0,3 . 22,4= 6,72 l\)

b)

Theo PTHH:

\(n_{HCl}= 3n_{Al}= 0,6 mol\)

\(\Rightarrow m_{HCl}= 0,6 . 36,5=21,9 g\)

\(\Rightarrow m_{dd HCl}= \dfrac{21,9 . 100}{15}= 146 g\) ( nếu ở tử là : 21,9 . 100% thì ở mẫu bạn chia cho 15% nhé)

a) \(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

Số mol : 2mol.........6mol

Sau p/ư : 0,2mol........ x

\(n_{HCl}=x=\dfrac{0,2.6}{2}=0,6\left(mol\right)\)

Ta có : \(n_{HCl}=\dfrac{16\%.m_{HCl}}{100\%.36,5}\)

\(\Rightarrow0,6=\dfrac{16.m_{HCl}}{100.36,5}\)

\(\Rightarrow m_{HCl}=\dfrac{2190}{16}=136,875\left(g\right)\)

b) \(n_{AlCl_3}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\)

\(m_{AlCl_3}=0,3.133,5=40,05\left(g\right)\)

\(\Rightarrow C_{\%ddAlCl_3}=\dfrac{40,05}{5,4+136,875}.100\approx27,95\%\)

nAl=5,4/27=0,2(mol)

PT:

2Al + 6HCl -> 2AlCl3 + 3H2↑

0.2........0.6........0.2..........0.3 (mol)

a) mHCl=0,6.36,5=21,9 (g)

b)m_AlCl3=0.2*133.5=26.7(g)

=> C% (AlCl₃)=(26.7/(21.9*100/16+5.4))*100%=18.766%

n_Al = 0,2 mol

2Al + 6HCl \(\rightarrow\) 2AlCl₃ + 3H₂

0,2.....0,6...........0,6..........0,3

\(\Rightarrow\) m_{HCl đã dùng} = \(\dfrac{0,6.36,5.100}{16}\) = 136,875 (g)

\(\Rightarrow\) m_{dd sau phản ứng} = 5,4 + 136,875 - ( 0,3.2 ) = 141,675 (g)

\(\Rightarrow\) C%_{dd muối sau phản ứng} = \(\dfrac{0,2.133,5.100}{141,675}\) \(\approx\) 18,8%

\(n_{MgCO_3}=\dfrac{21}{84}=0,25\left(mol\right)\\ a.MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\\ n_{HCl}=0,25.2=0,5\left(mol\right)\\ b.V_{ddHCl}=\dfrac{0,5}{2}=0,25\left(l\right)=250\left(ml\right)\)

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ Mg+2HCl\to MgCl_2+H_2\\ MgO+2HCl\to MgCl_2+H_2O\\ \Rightarrow n_{Mg}=0,1(mol)\\ \Rightarrow \%_{Mg}=\dfrac{0,1.24}{6,4}.100\%=37,5\%\\ \Rightarrow \%_{MgO}=100\%-37,5\%=62,5\%\)

\(b,n_{MgO}=\dfrac{6,4-0,1.24}{40}=0,1(mol)\\ \Rightarrow n_{HCl}=2.0,1+2.0,1=0,4(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,4}{0,5}=0,8(l)\\ c,n_{MgCl_2}=0,1+0,1=0,2(mol)\\ \Rightarrow C_{M_{MgCl_2}}=\dfrac{0,2}{0,8}=0,25M\)

Đáp án C

 m = m_oxit - m_{kim loại} = 25,1 - 15,3 = 9,8g

Bn tách câu hỏi ra nhỏ giúp mk !

a. \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

0,25 ..... 0,5 ................... 0,25 (mol) 

\(m_{Mg}=0,25.24=6\left(g\right)\)

\(\rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{6}{10}.100\%=60\%\\\%m_{MgO}=100\%-60\%=40\%\end{matrix}\right.\)

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

0,1 ....... 0,2 (mol) 

\(n_{HCl}=0,25+0,1=0,35\left(mol\right)\)

\(C_M\left(HCl\right)=\dfrac{0,35}{0,1}=3,5\left(M\right)\)