\(n_{CaCO_3}=0,2\left(mol\right)\)

\(n_{HCl}=0,6\left(mol\right)\)

\(2HCl+CaCO_3-->CaCl_2+CO_2\uparrow+H_2O\)

\(\dfrac{0,6}{2}>\dfrac{0,2}{1}\) => HCl dư

\(m_{dd1}=100+20-0,2.44=111,2\left(g\right)\)

\(MgCO_3+2HCl-->MgCl_2+CO_2\uparrow+H_2O\)

x.....................2x..................x................x.............x

\(m_{dd2}=84x+100-44x=111,2\)

=> x=0,28

\(m_{MgCO_3}=0,28.84=23,52\left(g\right)\)

cho tác dụng vs nước

tan tạo dd đục CaO

tan K2O , P2O5

P2O5 và K2O ( sp : H3PO4, KOH)

dùng quỳ tím

xanh là K2O đỏ là P2O5

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cách nhận bt là v t hay dùng sơ đồ nên k vt = chữ đc bạn dựa vào đó viết nha

\(n_{CH_4}=\dfrac{16}{16}=1\left(mol\right)\)

\(CH_4+2O_2-t^0->CO_2+2H_2O\)

1......................................1

\(V_{CO_2}=1.22,4=22,4\left(l\right)\)

\(\left\{{}\begin{matrix}e=p\\e+p+n=40\\e+p-n=12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}e=p=13\\n=14\end{matrix}\right.\)

X : Nhôm(Al)

\(CaCO_3+2HCl-->CaCl_2+CO_2\uparrow+H_2O\)

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

\(CaCO_3+2HCl-->CaCl_2+CO_2\uparrow+H_2O\)

x....................2x

\(MgCO_3+2HCl-->MgCl_2+CO_2\uparrow+H_2O\)

y....................2y

\(\left\{{}\begin{matrix}2x+2y=0,1\\100x+84y=4,68\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,03\\y=0,02\end{matrix}\right.\)

\(m_{CaCO_3}=100.0,03=3\left(g\right)\)

\(m_{MgCO_3}=4,68-3=1,68\left(g\right)\)

\(\%CaCO_3=\dfrac{3}{4,68}.100\%\approx64,1\%\)

\(\%MgCO_3=100\%-64,1\%=35,9\%\)

Gọi CTC là FexOy

\(n_{Fe}=0,1\left(mol\right)\)

\(Fe_xO_y+yCO-t^0->xFe+yCO_2\uparrow\)

0,1/x.....................................0,1

\(M_{Fe_xO_y}=56x+16y=\dfrac{8}{\dfrac{0,1}{x}}\)

\(\Leftrightarrow56x+16y=80x\)

\(\Leftrightarrow16y=24x\)

\(\Leftrightarrow\dfrac{x}{y}=\dfrac{16}{24}=\dfrac{2}{3}\Rightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)

CTHH : Fe2O3

\(n_{H2}=0,4\left(mol\right)\)

\(2Al+3H_2SO4-->Al2\left(SO4\right)_3+3H_2\uparrow\)

x...........1,5x.................................................1,5x

\(Fe+H2SO4-->FeSO4+H2\uparrow\)

y............y..............................................y

\(\left\{{}\begin{matrix}1,5x+y=0,4\\27x+56y=11\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

\(m_{Al}=0,2.27=5,4\left(g\right)\)

\(m_{Fe}=11-5,4=5,6\left(g\right)\)

\(V_{ddH2SO4}=\dfrac{0,2.1,5+0,1}{2}=0,2\left(l\right)\)

bổ sung :

2Na+2H2O-->2NaOH+H2

2Al+2NaOH+2H2O--->2NaAlO2+3H2