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a) Các PTHH xảy ra:
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(0,1--0,2------0,1\)
\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)
\(0,1---0,2\)
b) \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,1\cdot24=2,4\left(g\right)\)
\(\Rightarrow m_{ZnO}=10,5-2,4=8,1\left(g\right)\)
c) \(n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\)
\(\Rightarrow n_{HCl}=0,2+0,2=0,4\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,4\cdot36,5=14,6\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{14,6\cdot100}{20}=73\left(g\right)\)
\(\Rightarrow V_{ddHCl}=\dfrac{73}{1,1}\approx66,4\left(cm^3\right)\)
a. PTHH:
2Al + 3H2SO4 ---> Al2(SO4)3 + 3H2
Cu + H2SO4 ---x--->
b. Theo PT: \(n_{Al}=\dfrac{2}{3}.n_{H_2}=\dfrac{2}{3}.\dfrac{6,72}{22,4}=0,2\left(mol\right)\)
=> \(m_{Al}=0,2.27=5,4\left(g\right)\)
=> \(m_{Cu}=10-5,4=4,6\left(g\right)\)
c. \(\%_{m_{Al}}=\dfrac{5,4}{10}.100\%=54\%\)
\(\%_{m_{Cu}}=100\%-54\%=46\%\)
d. Theo PT: \(n_{H_2SO_4}=n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
=> \(m_{H_2SO_4}=0,3.98=29,4\left(g\right)\)
Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{29,4}{m_{dd_{H_2SO_4}}}.100\%=20\%\)
=> \(m_{dd_{H_2SO_4}}=147\left(g\right)\)
a) PTHH : \(Fe+2HCl-t^o->FeCl_2+H_2\) (1)
\(2Fe+3Cl_2-t^o->2FeCl_3\) (2)
\(Cu+Cl_2-t^o->CuCl_2\) (3)
b) Theo pthh (1) : \(n_{Fe}=n_{H2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)
Theo pthh (2) và (3) : \(\Sigma n_{Cl2}=\dfrac{3}{2}n_{Fe}+n_{Cu}\)
\(\Rightarrow\dfrac{6,72}{22,4}=\dfrac{3}{2}.0,1+n_{Cu}\)
\(\Rightarrow0,3=0,15+n_{Cu}\)
\(\Rightarrow n_{Cu}=0,15\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,15.64=9,6\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{5,6}{5,6+9,6}\cdot100\%\approx36,84\%\\\%m_{Cu}=100\%-36,84\%=63,16\%\end{matrix}\right.\)
a) PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\) (1)
\(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\) (2)
b) Ta có: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\\Sigma n_{H_2SO_4}=0,2\cdot1=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{Zn}=n_{ZnO}=n_{H_2SO_4\left(1\right)}=n_{H_2SO_4\left(2\right)}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Zn}=0,1\cdot65=6,5\left(g\right)\\m_{ZnO}=0,1\cdot81=8,1\left(g\right)\end{matrix}\right.\)
c) Theo các PTHH: \(\left\{{}\begin{matrix}n_{Zn}=n_{ZnSO_4\left(1\right)}=0,1mol\\n_{ZnO}=n_{ZnSO_4\left(2\right)}=0,1mol\end{matrix}\right.\)
\(\Rightarrow\Sigma n_{ZnSO_4}=0,2mol\) \(\Rightarrow C_{M_{ZnSO_4}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
(Coi như thể tích dd thay đổi không đáng kể)
PTHH\(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\)
tl............1................2.............2.............1.............1..(mol
br 0,1.................0,2......................................0,1(mol)
NaCl không phản ứng đc vsHCl
b)\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(C_{MHCl}=\dfrac{0,2}{0,4}=0,5\left(M\right)\)(đổi 400ml=0,4(l))
c)\(Tacom_{Na_2CO_3}=0,1.106=10,6\left(g\right)\)
\(\Rightarrow\%m_{Na_2CO_3}=\dfrac{10,6}{20}.100=53\%\)
\(\Rightarrow\%mNaCl=100\%-53\%=47\%\)
a) PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\) (1)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\) (2)
b) Ta có: \(\Sigma n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Gọi số mol của Mg là \(a\) \(\Rightarrow n_{H_2\left(1\right)}=a\)
Gọi số mol của Zn là b \(\Rightarrow n_{H_2\left(2\right)}=b\)
Ta lập được hệ phương trình:
\(\left\{{}\begin{matrix}a+b=0,4\\24a+65b=17,8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,2\cdot24}{17,8}\cdot100\%\approx26,97\%\\\%m_{Zn}\approx73,03\%\end{matrix}\right.\)
a) $Zn + 2HCl \to ZnCl_2 + H_2$
b) $n_{Zn} = n_{H_2} = \dfrac{2,24}{22,4} = 0,1(mol)$
$\%m_{Zn} = \dfrac{0,1.65}{15}.100\% = 43,33\%$