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CuO +H2-->Cu+H2O
x(80x)------x(64x)
Fe2O3+3H2--->2Fe+3H2O
160y--------------112y
Theo bài ra ta có hệ pt
\(\left\{{}\begin{matrix}80x+160y=24\\64x+112y=17,6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
%m\(_{CuO}=\frac{80.0,1}{24}.100\%=33,33\%\)
%m\(_{Fe2O3}=100-33,33=66,67\%\)
\(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\)
x___________________ x______________
0,1_______________0,2______________mol
\(NaHCO_3+HCl\rightarrow NaCl+CO_2+H_2O\)
y____________________ y_________________
0,1___________________0,1__________ mol
\(n_{CO2}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}106x+84y=0\\x+y=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
\(m_{Na2CO3}=0,1.106=10,6\left(g\right)\)
\(m_{NaHCO3}=19-10,6=8,4\left(g\right)\)
\(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\)
\(NaHCO_3+HCl\rightarrow NaCl+CO_2+H_2O\)
Gọi \(\left\{{}\begin{matrix}n_{Na2CO3}=x\left(mol\right)\\n_{NaHCO3}:y\left(mol\right)\end{matrix}\right.\)
Giải hệ PT:
\(\left\{{}\begin{matrix}106a+84b=19\\a+b=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Na2CO3}=10,6\left(g\right)\\m_{NaHCO3}=8,4\left(g\right)\end{matrix}\right.\)
2Al+3H2SO4--->Al2(SO4)3+3H2
x------1,5x
Mg+H2SO4----->MgSO4+H2
y-----y
n\(_{H2SO4}=\frac{39,2}{98}=0,4\left(mol\right)\)
Theo bài ra ta có pt
\(\left\{{}\begin{matrix}27x+24y=7,8\\1,5x+y=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
%m\(_{Al}=\frac{0,2.27}{7,8}.100\%=69,23\%\)
%m\(_{Mg}=100-69,23=30,77\%\)
Đặt :
nNa2CO3 = x mol
nNaHCO3 = y mol
<=> 106x + 84y = 50 (1)
2NaHCO3 -to-> Na2CO3 + CO2 + H2O
y_______________0.5y
mNa2CO3 = 106x + 53y = 34.5 (2)
Giải (1) và (2) :
x = 4/53
y = 0.5
mM = 4/53 * 106 + 0.5*84 = 50g
\(n_{Na_2CO_3}=x;n_{NaHCO_3}=y\)
Na2CO3 không bị phân hủy
\(PTHH:2NaHCO_3\underrightarrow{t^o}Na_2CO_3+H_2O+CO_2\uparrow\)
(mol) 2 1
(mol) y y/2
Ta có: \(\left\{{}\begin{matrix}106x+84y=50\\106x+\frac{106y}{2}=34,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{4}{53}\\y=0,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n_{Na_2CO_3}=\frac{4}{53}\left(mol\right)\rightarrow m_{Na_2CO_3}=\frac{4}{53}.106=8\left(g\right)\\n_{NaHCO_3}=0,5\left(mol\right)\rightarrow m_{NaHCO_3}=0,5.84=42\left(g\right)\end{matrix}\right.\)