\(n_{CuO}=\dfrac{10}{80}=0,125\left(mol\right)\)

\(CuO+2HCl\rightarrow CuCl_2+H_2O\)

\(\Rightarrow n_{HCl}=2n_{CuO}=0,25\left(mol\right)\)

\(\Rightarrow V_{dd}=\dfrac{0,25}{3}=\dfrac{1}{12}\left(l\right)\)

\(n_{CuCl_2}=n_{CuO}=0,125\left(mol\right)\)

\(\Rightarrow C_{M\left(CuCl_2\right)}=\dfrac{0,125}{\dfrac{1}{12}}=1,5M\)

cảm ơn bạn

- Chất rắn không tan là Cu

CuO + 2HCl \(\rightarrow\) CuCl₂ + H₂O

m_Cu = 9,6 (g)

\(\Rightarrow\) m_CuO = 17,6 - 9,6 = 8 (g)

a) \(n_{MgCO_3}=\dfrac{16,8}{84}=0,2\left(mol\right)\)

PTHH: MgCO₃ + 2HCl ---> MgCl₂ + CO₂ + H₂O

            0,2--------------------->0,2----->0,2

=> \(m=m_{MgCl_2}=0,2.95=19\left(g\right)\)

\(m_{\text{dd}.sau.p\text{ư}}=150+16,8-0,2.44=158\left(g\right)\)

=> \(C\%_{MgCl_2}=\dfrac{19}{158}.100\%=12,025\%\)

b) CO₂ + Ca(OH)₂ ---> CaCO₃ + H₂O

     0,2----->0,2------------>0,2

=> \(\left\{{}\begin{matrix}m_{kt}=m_{CaCO_3}=0,2.100=20\left(g\right)\\V_{\text{dd}Ca\left(OH\right)_2}=\dfrac{0,2}{3}=\dfrac{1}{15}\left(l\right)\end{matrix}\right.\)

1.

\(m_{HCl}=\dfrac{10,95.75}{100}=8,2125\left(g\right)\)

\(\Rightarrow n_{HCl}=\dfrac{8,2125}{35,5}=0,225\left(mol\right)\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(\Rightarrow n_{Fe_2O_3}=\dfrac{1}{6}n_{HCl}=0,0375\left(mol\right)\)

\(n_{FeCl_3}=\dfrac{1}{3}n_{HCl}=0,075\left(mol\right)\)

\(\Rightarrow C\%=\dfrac{0,075.162,5}{0,0375.160+75}.100\%=15,05\%\)

cảm ơn bạn 

a. \(m_{Al.pứ}=15-9,6=5,4\left(g\right)\)

b. \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{5,4.100\%}{15}=36\%\\\%m_{Cu}=\dfrac{9,6.100\%}{15}=64\%\end{matrix}\right.\)

c. \(n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.\dfrac{5,4}{27}=0,3\left(mol\right)\Rightarrow V_{H_2\left(đkc\right)}=0,3.24,79=7,437\left(l\right)\)

d. \(\%m_{AlCl_3}=\dfrac{0,2.133,5.100\%}{15+200-9,6}=13\%\)

Ba+2HCl->BaCl2+H2

2Al+6HCl->2AlCl3+3H2

=>X gồm BaCl2 ,AlCl3

a) \(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: CaCO₃ + 2HCl ---> CaCl₂ + CO₂ + H₂O

             0,3<-----0,6<------------------0,3

=> m = 0,3.100 = 30 (g)

b) \(C\%_{HCl}=\dfrac{0,6.36,5}{150}.100\%=14,6\%\)

nCuO = 6.4/80=0.08 mol

mHCl = 7.3 g

nHCl = 7.3/36.5=0.2 mol

CuO + 2HCl --> CuCl2 + H2O

Bđ: 0.08___0.2

Pư : 0.08__0.16_____0.08

Kt : 0_____0.04_____0.08

mHCl dư = 0.04*36.5 = 1.46 g

mCuCl2 = 0.08*135=10.8 g

mdd sau phản ứng = 6.4 + 36.5 = 42.9 g

C%HCl dư = 1.46/42.9*100% = 3.4%

C%CuCl2 = 10.8/42.9*100% = 25.17%