a) Gọi số mol Ca, CaCO₃ là a, b (mol)

=> 40a + 100b = 19 (1)

\(m_{HCl}=\dfrac{500.4,38}{100}=21,9\left(g\right)\)

PTHH: Ca + 2HCl --> CaCl₂ + H₂

             a--->2a------->a----->a

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

                b------>2b------>b------>b

=> \(\overline{M}_Y=\dfrac{2a+44b}{a+b}=13,6.2=27,2\left(g/mol\right)\)

=> 25,2a =  16,8b (2)

(1)(2) => a = 0,1 (mol); b = 0,15 (mol)

\(\left\{{}\begin{matrix}m_{Ca}=0,1.40=4\left(g\right)\\m_{CaCO_3}=0,15.100=15\left(g\right)\end{matrix}\right.\)

b) 

m_{dd sau pư} = 19 + 500 - 0,1.2 - 0,15.44 = 512,2 (g)

m_HCl(dư) = 21,9 - 36,5(2a + 2b) = 3,65 (g)

m_CaCl2 = 111(a + b) = 27,75 (g)

\(\left\{{}\begin{matrix}C\%_{CaCl_2}=\dfrac{27,75}{512,2}.100\%=5,418\%\\C\%_{HCl\left(dư\right)}=\dfrac{3,65}{512,2}.100\%=0,713\%\end{matrix}\right.\)

a) Gọi số mol Ca, CaCO₃ là a, b (mol)

=> 40a + 100b = 19 (1)

\(m_{HCl}=\dfrac{500.4,38}{100}=21,9\left(g\right)\)

PTHH: Ca + 2HCl --> CaCl₂ + H₂

             a--->2a------->a----->a

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

                b------>2b------>b------>b

=> \(\overline{M}_Y=\dfrac{2a+44b}{a+b}=13,6.2=27,2\left(g/mol\right)\)

=> 25,2a =  16,8b (2)

(1)(2) => a = 0,1 (mol); b = 0,15 (mol)

\(\left\{{}\begin{matrix}m_{Ca}=0,1.40=4\left(g\right)\\m_{CaCO_3}=0,15.100=15\left(g\right)\end{matrix}\right.\)

b) 

m_{dd sau pư} = 19 + 500 - 0,1.2 - 0,15.44 = 512,2 (g)

m_HCl(dư) = 21,9 - 36,5(2a + 2b) = 3,65 (g)

m_CaCl2 = 111(a + b) = 27,75 (g)

\(\left\{{}\begin{matrix}C\%_{CaCl_2}=\dfrac{27,75}{512,2}.100\%=5,418\%\\C\%_{HCl\left(dư\right)}=\dfrac{3,65}{512,2}.100\%=0,713\%\end{matrix}\right.\)

Mg+2HCl->MgCl2+H2

x------2x--------x---------x

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

y---------3y-----y--------3\2y

ta có :

\(\left\{{}\begin{matrix}24x+27y=11,7\\x+\dfrac{3}{2}y=0,6\end{matrix}\right.\)

=>x=0,15 mol, y=0,3 mol

=>%mMg=\(\dfrac{0,15.24}{11,7}.100=30,77\%\)

=>%mAl=100-30,77=69,23%

b)

m HCl=1,2.36,5=43,8g

=>C%=\(\dfrac{43,8}{200}.100\)=21,9%

\(a,Đặt:n_{Mg}=g\left(mol\right);n_{Fe}=j\left(mol\right)\left(g,j>0\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ \Rightarrow\left\{{}\begin{matrix}24g=56j=9,2\\22,4g+22,4j=5,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}g=0,15\\j=0,1\end{matrix}\right.\\ \Rightarrow\%m_{Mg}=\dfrac{0,15.24}{9,2}.100\approx39,13\%\\ \Rightarrow\%m_{Fe}\approx60,87\%\\ b,n_{HCl}=2n_{H_2}=\dfrac{2.5,6}{22,4}=0,5\left(mol\right)\\ \Rightarrow C_{MddHCl}=\dfrac{0,5}{0,2}=2,5\left(M\right)\\ c,m_{muối}=m_{FeCl_2}+m_{MgCl_2}=95g+127j=95.0,15+127.0,1=26,95\left(g\right)\)

Fe+2HCl->FeCl₂+H₂

x-----------------------x mol

Mg+2HCl->MgCl₂+H₂

y-------------------------y mol

ta có\(\left\{{}\begin{matrix}56x+24y=9,2\\x+y=0,25\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=0,1\\y=0,15\end{matrix}\right.\)

=>%m_Fe=\(\dfrac{0,1.56}{9,2}.100\)=60,87%

=>%m _Mg=39,13%

Ta có : n _HCl=0,1.2+0,15.2=0,5 mol

=>CM_HCl=\(\dfrac{0,5}{0,2}\)=2,5M

=>m _muối =0,1.127+0,15.95=26,95g

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Đáp án B

Dùng phương pháp đường chéo tính được mol CO₂ là 0,2 mol và H₂ là 0,15 mol.