Cho 30g hỗn hợp gồm Cu và Zn tác dụng hết với 200g dung dịch HCl thu được 5600ml khí (đktc).a.Tính khối lượng mỗi kim loại.b. Tính thành phần % khối lượng mỗi kim loại theo khối lượng.c.Tính nồng độ % HCl.
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4 tháng 4 2020
a,
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(n_{H2}=\frac{5,6}{22,4}=0,25\left(mol\right)\)
\(n_{Zn}=n_{H2}=0,25\left(mol\right)\)
\(m_{Zn}=0,25.65=16,25\left(g\right)\)
\(m_{Cu}=30-16,25=13,75\left(g\right)\)
b)
\(\%m_{Zn}=\frac{16,25}{30}.100\%=54,17\%\)
\(\%m_{Cu}=100\%-54,17\%=45,83\%\)
c)
\(n_{HCl}=2n_{H2}=0,5\left(mol\right)\)
\(\Rightarrow CM_{HCl}=\frac{0,5.36,5}{200}.100\%=9,125\%\)
4 tháng 1 2023
a, PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Ta có: \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
Theo PT: \(n_{Fe}=n_{H_2}=0,25\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,25.56=14\left(g\right)\)
mCu = 20,4 - 14 = 6,4 (g)
b, \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{14}{20,4}.100\%\approx68,63\%\\\%m_{Cu}\approx31,37\%\end{matrix}\right.\)
c, Theo PT: \(n_{HCl}=2n_{H_2}=0,5\left(mol\right)\)
\(\Rightarrow C\%_{HCl}=\dfrac{0,5.36,5}{200}.100\%=9,125\%\)
MN
6 tháng 12 2021
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{Fe}=n_{H_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)
\(\Rightarrow m_{Fe}=0.15\cdot56=8.4\left(g\right)\)
\(m_{Cu}=m_{hh}-m_{Fe}=15-8.4=6.6\left(g\right)\)
\(n_{HCl}=2n_{H_2}=0.15\cdot2=0.3\left(mol\right)\)
\(C_{M_{HCl}}=\dfrac{0.3}{0.2}=1.5\left(M\right)\)
26 tháng 3 2022
Gọi \(\left\{{}\begin{matrix}n_{Ag}=a\left(mol\right)\\n_{FeO}=b\left(mol\right)\end{matrix}\right.\)
\(n_{SO_2}=\dfrac{1,344}{22,4}=0,6\left(mol\right)\)
PTHH:
\(2Ag+2H_2SO_4\rightarrow Ag_2SO_4+SO_2\uparrow+2H_2O\)
a a \(\dfrac{a}{2}\) \(\dfrac{a}{2}\)
\(2FeO+4H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+SO_2\uparrow+4H_2O\)
b 2b \(\dfrac{b}{2}\) \(\dfrac{b}{2}\)
Hệ pt
\(\left\{{}\begin{matrix}108a+72b=11,52\\\dfrac{a}{2}+\dfrac{b}{2}=0,06\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,08\left(mol\right)\\b=0,04\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{Ag}=0,08.108=8,64\left(g\right)\\m_{FeO}=0,04.72=2,88\left(g\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\%m_{Ag}=\dfrac{8,64}{11,52}=75\%\\\%m_{FeO}=100\%-75\%=25\%\end{matrix}\right.\)
b, \(\rightarrow n_{H_2SO_4}=0,08+0,4.2=0,16\left(mol\right)\\ \rightarrow C_{MddH_2SO_4}=\dfrac{0,16}{0,8}=0,2M\)
c, \(n_{NaOH}=1,25.0,5=0,625\left(mol\right)\)
PTHH:
\(6NaOH+Fe_2\left(SO_4\right)_3\rightarrow2Fe\left(OH\right)_3+3Na_2SO_4\)
LTL: \(\dfrac{0,625}{6}>\dfrac{0,04}{2}\) => NaOH dư
Theo pthh:
\(\left\{{}\begin{matrix}n_{NaOH\left(pư\right)}=6n_{Fe_2\left(SO_4\right)_3}=6.0,04=0,24\left(mol\right)\\n_{Na_2SO_4}=3n_{Fe_2\left(SO_4\right)_3}=3.0,04=0,12\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}C_{MddNaOH\left(dư\right)}=\dfrac{0,24}{0,5}=0,48M\\C_{MddNa_2SO_4}=\dfrac{0,12}{0,5}=0,24M\end{matrix}\right.\)
12 tháng 2 2020
1.
a )\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(n_{H2}=\frac{5,6}{22,4}=0,25\left(mol\right)\)
\(n_{Zn}=n_{H2}=0,25\left(mol\right)\)
\(\rightarrow,m_{Zn}=0,25.65=16,25\left(g\right);m_{Cu}=30-16,25=13,75\left(g\right)\)
b)
\(\%m_{Zn}=\frac{16,25}{30}.100\%=54,17\%\)
\(\%m_{Cu}=100\%-54,17\%=45,83\%\)
c)
\(n_{HCl}=2n_{H2}=0,5\left(mol\right)\)
\(C\%_{HCl}=\frac{0,5.36,5}{200}.100\%=9,125\%\)
2.
a)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{H2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)\(m_{Fe}=0,15.65=8,4\left(g\right),m_{Ag}=15-8,4=6,6\left(g\right)\)
b)
\(\%m_{Fe}=\frac{8,4}{15}.100\%=56\%\)
\(\%m_{Ag}=100\%-56\%=44\%\)
c)
\(n_{HCl}=2n_{H2}=0,3\left(mol\right)\)
\(\rightarrow m_{dd_{HCL}}=\frac{0,3.36,5}{15,6\%}=70,19\left(g\right)\)
21 tháng 3 2023
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, Ta có: \(n_{H_2}=\dfrac{22,4}{22,4}=1\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2}=1\left(mol\right)\Rightarrow m_{Zn}=1.65=65\left(g\right)\)
\(\Rightarrow m_{Cu}=80,5-65=15,5\left(g\right)\)
a, \(n_{H_2}=0,25\left(mol\right)\)
Bảo toàn e:
\(2n_{Zn}=2n_{H_2}\Rightarrow n_{Zn}=0,25\left(mol\right)\)
\(\Rightarrow m_{Zn}=16,25\left(g\right)\)
\(\Rightarrow m_{Cu}=13,75\left(g\right)\)
b, \(\%m_{Cu}=\dfrac{13,75}{30}=45,83\%\)
\(\Rightarrow\%m_{Zn}=100\%-45,83\%=54,17\%\)