mHCl= 3,65%.400= 14,6(g) => nHCl=14,6/36,5=0,4(mol)

a) PTHH: Mg +2 HCl -> MgCl2 + H2

0,2__________0,4_____0,2____0,2(mol)

V(H2,đktc)=0,2.22,4=4,48(l)

b)mMg=0,2.24=4,8(g)

c) mMgCl2= 0,2.95=19(g)

mddMgCl2= 400+4,8 - 0,2.2= 404,4(g)

=> C%ddMgCl2= (19/404,4).100=4,698%

\(n_{HCl}=\dfrac{400\cdot3.65\%}{36.5}=0.4\left(mol\right)\)

\(C_{M_{HCl}}=\dfrac{0.4}{2}=0.2\left(M\right)\)

\(m_{ct}=\dfrac{3,65.400}{100}=14,6\left(g\right)\)

\(n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)

Pt : \(2HCl+Ba\left(OH\right)_2\rightarrow BaCl_2+2H_2O|\)

           2              1                 1            2

         0,4            0,2              0,2

a) \(n_{Ba\left(OH\right)2}=\dfrac{0,4.1}{2}=0,2\left(mol\right)\)

\(m_{Ba\left(OH\right)2}=0,2.171=34,2\left(g\right)\)

\(m_{ddBa\left(OH\right)2}=\dfrac{34,2.100}{17,1}=200\left(g\right)\)

b) \(n_{BaCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{BaCl2}=0,2.208=41,6\left(g\right)\)

\(m_{ddspu}=400+200=600\left(g\right)\)

\(C_{BaCl2}=\dfrac{41,6.100}{600}=6,93\)⁰/₀

 Chúc bạn học tốt

\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

a          2a          a            a

\(FeS+2HCl\rightarrow FeCl_2+H_2S\uparrow\)

b            2b            b            b

\(n_{HCl}=\dfrac{400\times7.3}{100\times36.5}=0.8mol\)

\(n_X=\dfrac{4.48}{22.4}=0.2mol\)

\(M_X=2\times9=18\Leftrightarrow\dfrac{2a+34b}{a+b}=18\)

Ta có: \(\left\{{}\begin{matrix}a+b=0.2\\\dfrac{2a+34b}{a+b}=18\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a+b=0.2\\2a+34b=3.6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=0.1\\b=0.1\end{matrix}\right.\)

a. \(\%V_{H_2}=\dfrac{0.1\times22.4\times100}{4.48}=50\%\)

\(\%V_{H_2S}=100-50=50\%\)

b. \(a=0.1\times56+0.1\times88=14.4g\)

\(\%m_{Fe}=\dfrac{0.1\times56}{14.4}\times100=38.8\%\)

\(\%m_{FeS}=100-38.8=61.2\%\)

c. m dung dịch sau phản ứng\(=14.4+400-0.1\times2-0.1\times34=410.8g\)

nHCl phản ứng\(=2\times0.1+2\times0.1=0.4mol\)

nHCl dư = 0.8 - 0.4 = 0.4 mol

\(C\%_{HCldu}=\dfrac{0.4\times36.5\times100}{410.8}=3.55\%\)

\(C\%_{FeCl_2}=\dfrac{0.2\times127\times100}{410.8}=6.18\%\)

giúp vs

 ko phải là 10g M CO3 mà là 10g MCO3

a)

$MCO_3 + 2HCl \to MCl_2 + CO_2 + H_2O$
Theo PTHH : 

$n_{H_2O} = n_{CO_2} = a(mol)$
$n_{HCl} = 2n_{CO_2} = 2a(mol)$

Bảo toàn khối lượng : 

$10 + 2a.36,5 = 44a + 18a + 11,1 \Rightarrow a = 0,1$
$V = 0,1.22,4 = 2,24(lít)$

b) $n_{HCl} = 2a = 0,2(mol)$
$m = \dfrac{0,2.36,5}{3,65\%} = 200(gam)$
c)
$m_{dd} = 10 + 200 - 0,1.44 = 205,6(gam)$

$C\%_{muối} = \dfrac{11,1}{205,6}.100\% = 5,4\%$

d)

$M_{MCO_3} = \dfrac{10}{0,1} = 100 \Rightarrow M = 40(Canxi)$

\(a)n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1       0,2          0,1           0,1
\(n_{Fe_2O_3}=\dfrac{21,6-56.0,1}{160}=0,1mol\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2\)
0,1            0,6            0,2              0,3
\(V_{ddHCl}=\dfrac{0,2+0,6}{1}=0,8l\\ b.C_{M_{FeCl_2}}=\dfrac{0,1}{0,8}=0,125M\\ C_{M_{FeCl_3}}=\dfrac{0,2}{0,8}=0,25M\)

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

0,1       0,2        0,1            0,1

\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)

=>\(n_{H_2}=0.1\left(mol\right)\)

=>\(n_{HCl}=0.2\left(mol\right)\)

\(V_{HCl}=\dfrac{0.2}{4}=0.05\left(lít\right)\)

b: \(V_{H_2}=0.1\cdot22.4=2.24\left(lít\right)\)