\(a)n_{Fe}=\dfrac{11,2}{56}=0,2mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ 0,2\rightarrow0,4-\rightarrow0,2-\rightarrow0,2\)

\(V_{H_2}=0,2.22,4=4,48l\\ b)m_{FeCl_2}=0,2.127=25,4g\\ c)C_{M_{HCl}}=\dfrac{0,4}{0,1}=4M\)

a)

$n_{CO_2} = \dfrac{0,224}{22,4} = 0,01(mol)$
\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

0,01             0,02                        0,01                            (mol)

$C_{M_{HCl}} = \dfrac{0,02}{0,1} = 0,2M$

b)

$\%m_{CaCO_3} = \dfrac{0,01.100}{1,5}.100\% = 66,67\%$

$\%m_{CaSO_4} = 100\% -66,67\% = 33,33\%$

a/ Fe + 2HCl \(\rightarrow\) FeCl₂ + H₂

n_H2 = \(\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PTHH: n_H2 = n_Fe = 0,15 (mol) \(\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)

\(\Rightarrow m_{Cu}=11-8,4=2,6\left(g\right)\)

\(\Rightarrow\%m_{Fe}=\dfrac{8,4}{11}.100\%\approx76,4\%\)

\(\Rightarrow\%m_{Cu}=100-76,4\approx23,6\%\)

b/ Theo PTHH ta có: n_HCl = 2n_Fe = 2.0,15 = 0,3 (mol)

\(\Rightarrow V_{ddHCl}=\dfrac{0,3}{2}=0,15\left(M\right)\)

c/ m_HCl = 36,5 . 0,3 = 10,95(g)

\(\Rightarrow C\%_{HCl}=\dfrac{m_{HCl}}{m_{ddHCl}}.100\%=\dfrac{10,95}{200}.100\%=5,475\%\)

a, PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PT: \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,15.56}{11}.100\%\approx76,36\%\\\%m_{Cu}\approx23,64\%\end{matrix}\right.\)

b, Theo PT: \(n_{HCl}=2n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow V_{ddHCl}=\dfrac{0,3}{2}=0,15\left(l\right)\)

c, \(C\%_{HCl}=\dfrac{0,3.36,5}{200}.100\%=5,475\%\)

PTHH\(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\)

tl............1................2.............2.............1.............1..(mol

br     0,1.................0,2......................................0,1(mol)

NaCl không phản ứng đc vsHCl

b)\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(C_{MHCl}=\dfrac{0,2}{0,4}=0,5\left(M\right)\)(đổi 400ml=0,4(l))

c)\(Tacom_{Na_2CO_3}=0,1.106=10,6\left(g\right)\)

\(\Rightarrow\%m_{Na_2CO_3}=\dfrac{10,6}{20}.100=53\%\)

\(\Rightarrow\%mNaCl=100\%-53\%=47\%\)

\(a)n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ 0,2\leftarrow-0,3\leftarrow-0,1\leftarrow---0,3\)

\(a=m_{Al}=0,2.27=5,4g\\ b)m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2g\\ c)C_{\%H_2SO_4}=\dfrac{0,3.98}{100}\cdot100=29,4\%\)

a)nH2​​=22,46,72​=0,3mol2Al+3H2​SO4​→Al2​(SO4​)3​+3H2​0,2←−0,3←−0,1←−−−0,3

$a=m^{Al}=0,2.27=5,4gb)m^{A{l}^2{\left(S{O}^4\right)}^3}=0,1.342=34,2gc)C^{\%H^{2}S{O}^4}=\genfrac{}{}{0ex}{}{0,3.98}{100}\cdot 100=29,4\%$

chj kb vs em nha em ib rồi ạ