22
a
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
x--------------->x--------->x
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
y-------------->y--------->y
Có hệ: \(\left\{{}\begin{matrix}65x+56y=12,1\\136x+127y=26,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
\(\%m_{Zn}=\dfrac{0,1.65.100\%}{12,1}=53,72\%\\ \%m_{Fe}=\dfrac{0,1.56.100\%}{12,1}=46,28\%\)
b
\(V_{H_2}=\left(0,1+0,1\right).22,4=4,48\left(l\right)\)
23
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
x----------------------------->x
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
y-------------------------------->1,5y
Có hệ: \(\left\{{}\begin{matrix}24x+27y=3,75\\x+1,5y=\dfrac{3,92}{22,4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)
\(\%m_{Mg}=\dfrac{0,1.24.100\%}{3,75}=64\%\\ \%m_{Al}=\dfrac{0,05.27.100\%}{3,75}=36\%\)
\(30.\\ n_{Ba\left(OH\right)_2}=0,2.1=0,2mol \\ n_{BaCO_3}=\dfrac{29,55}{197}=0,15mol\\n_{BaCO_3}< n_{Ba\left(OH\right)_2}\\ TH1:tạo.BaCO_3\\ CO_2+Ba\left(OH\right)_2\rightarrow BaCO_3+H_2O\\ n_{CO_2}=n_{BaCO_3}=0,15mol\\ V_{CO_2}=0,15.22,4=3,36l\\ TH2:tạo.BaCO_3,Ba\left(HCO_3\right)_2\\ CO_2+Ba\left(OH\right)_2\rightarrow BaCO_3+H_2O\\ n_{CO_2}=n_{BaCO_3}=0,15mol\\ 2CO_2+Ba\left(OH\right)_2\rightarrow Ba\left(HCO_3\right)_2\\ n_{CO_2}=2.\left(0,2-0,15\right)=0,1\\ V_{CO_2}=\left(0,1+0,15\right)22,4=5,6l\)
\(29.\\ Fe_2O_3+3H_2\xrightarrow[]{t^0}2Fe+3H_2O\\ CuO+H_2\xrightarrow[]{t^0}Cu+H_2O\\ n_{Fe_2O_3}=a;n_{CuO}=b\\ \Rightarrow\left\{{}\begin{matrix}160a+80b=32\\56.2a+64b=24\end{matrix}\right.\\ \Rightarrow a=0,1,b=0,2\\ V=V_{H_2}=(0,1.3+0,2).22,4=11,2l\\m_{rắn}=m_{Cu}=24-0,1.2.56=12,8g\)
\(28.\\ n_{Fe_2O_3}=\dfrac{3,2}{160}=0,02mol\\ 3CO+Fe_2O_3\rightarrow2Fe+3CO_2\\ n_{CO}=n_{CO_2}=0,02.3=0,06mol\\ V_{CO}=0,06.22,4=1,344l\\ CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\\ n_{CaCO_3}=n_{CO_2}=0,06mol\\ m_{kt}=m_{CaCO_3}=0,06.100=6g\)
\(27.\)
Gọi hoá trị của kim loại M là n
\(4M+nO_2\xrightarrow[]{t^0}2R_2O_n\)
\(4M\) \(2\left(2M+16n\right)\)
1 1.667
\(\Leftrightarrow1.,667M\cdot4=4M+32n\)
\(\Leftrightarrow M=12n\)
Nếu n = 2 thì M là Mg
\(26.\\ n_{Na_2O}=\dfrac{6,2}{62}=0,1mol\\ V_{H_2O}=\dfrac{100}{1}=100ml\\ Na_2O+H_2O\rightarrow2NaOH\\ n_{NaOH}=0,1.2=0,2mol\\ C_{\%NaOH}=\dfrac{0,2.40}{6,2+100}\cdot100=7,5\%\)
\(C_M\) \(_{NaOH}=\dfrac{0,2}{0,1}=2M\)
\(25.\\ a.n_{HCl}=0,2.1=0,2mol\\ n_{NaOH}=0,5.0,3=0,15mol\\ NaOH+HCl\rightarrow NaCl+H_2O\\ \Rightarrow\dfrac{0,2}{1}>\dfrac{0,15}{1}\Rightarrow HCl.dư\)
⇒Quỳ tím hoá đỏ
\(NaOH+HCl\rightarrow NaCl+H_2O\)
0,15 0,15 0,15 0,15
\(V=0,2+0,3=0,5l\)
\(C_M\) \(_{NaOH}=\dfrac{0,15}{0,5}=0,3M\)
\(C_M\) \(_{HCl}=\dfrac{0,2-0,15}{0,5}=0,1M\)
\(24.\\ n_{Mg}=\dfrac{6}{24}=0,25mol\\ n_{H_2SO_4}=\dfrac{73,5.20}{100}:98=0,15mol\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ \Rightarrow\dfrac{0,25}{1}>\dfrac{0,15}{1}\Rightarrow Mg.dư\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
0,15 0,15 0,15 0,15
\(m_{dd}=0,15.24+73,5-0,15.2=76,8g\\ C_{\%MgSO_4}=\dfrac{0,15.120}{76,8}\cdot100=23,4375\%\)
\(23.\\ n_{H_2}=\dfrac{3,92}{22,4}=0,175mol\\ n_{Al}=a;n_{Mg}=b\\2 Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ \Rightarrow\left\{{}\begin{matrix}27a+24b=3,75\\1,5a+b=0,175\end{matrix}\right.\\ \Rightarrow a=0,05;b=0,1\\ \%m_{Al}=\dfrac{0,05.27}{3,75}\cdot100=36\%\\ \%m_{Mg}=100-36=64\%\)