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400ml = 0,4l
\(n_{HCl}=1.0,4=0,4\left(mol\right)\)
a) Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)
1 2 1 1
0,2 0,4 0,2
b) \(n_{Fe}=\dfrac{0,4.1}{2}=0,2\left(mol\right)\)
⇒ \(m_{Fe}=0,2.56=11,2\left(g\right)\)
\(n_{H2}=\dfrac{0,4.1}{2}=0,2\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)
c) \(n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)\)
Pt ; \(H_2+CuO\underrightarrow{t^o}Cu+H_2O|\)
1 1 1 1
0,2 0,3 0,2
Lập tỉ số so sánh : \(\dfrac{0,2}{1}< \dfrac{0,3}{1}\)
⇒ H2 phản ứng hết , CuO dư
⇒ Tính toán dựa vào số mol của H2
\(n_{Cu}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
⇒ \(m_{Cu}=0,2.64=12,8\left(g\right)\)
Chúc bạn học tốt
Em coi làm mấy bài hôm nay chưa làm nha, làm được nhiêu làm nè
\(n_{H2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
a) \(Fe+2HCl\rightarrow FeCl_2+H_2|\)
1 2 1 1
0,15 0,15 0,15
b) \(n_{Fe}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)
⇒ \(m_{Fe}=01,5.56=8,4\left(g\right)\)
c) \(n_{FeCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
⇒ \(m_{FeCl2}=0,2.127=25,4\left(g\right)\)
Chúc bạn học tốt
a, \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)
\(m_{HCl}=200.14,6\%=29,2\left(g\right)\Rightarrow n_{HCl}=\dfrac{29,2}{36,5}=0,8\left(mol\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Xét tỉ lệ: \(\dfrac{0,3}{1}< \dfrac{0,8}{2}\), ta được HCl dư.
Theo PT: \(n_{H_2}=n_{Zn}=0,3\left(mol\right)\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
b, \(n_{ZnCl_2}=n_{Zn}=0,3\left(mol\right)\Rightarrow m_{ZnCl_2}=0,3.136=40,8\left(g\right)\)
c, \(n_{HCl\left(pư\right)}=2n_{Zn}=0,6\left(mol\right)\Rightarrow n_{HCl\left(dư\right)}=0,2\left(mol\right)\)
Ta có: m dd sau pư = 19,5 + 200 - 0,3.2 = 218,9 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{HCl}=\dfrac{0,2.36,5}{218,9}.100\%\approx3,33\%\\C\%_{ZnCl_2}=\dfrac{40,8}{218,9}.100\%\approx18,64\%\end{matrix}\right.\)
\(a)n_{Zn}=\dfrac{19,5}{65}=0,3mol\\ n_{HCl}=\dfrac{200.14,6}{100.36,5}=0,8mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ \Rightarrow\dfrac{0,3}{1}< \dfrac{0,8}{2}\Rightarrow HCl.dư\\ n_{H_2}=n_{ZnCl_2}=n_{Zn}=0,3mol\\ V_{H_2}=0,3.22,4=6,72l\\ b)m_{ZnCl_2}=0,3.136=40,8g\\ c)n_{HCl.pư}=0,3.2=0,6mol\\ C_{\%ZnCl_2}=\dfrac{40,8}{200+19,5-0,3.2}\cdot100=18,64\%\\ C_{\%HCl.dư}=\dfrac{\left(0,8-0,6\right).36,5}{200+19,5-0,3.2}\cdot100=3,33\%\)
\(n_{HCl}=\dfrac{18.25}{36.5}=0.5\left(mol\right)\)
\(a.Mg+2HCl\rightarrow MgCl_2+H_2\)
\(b.\)
\(n_{Mg}=n_{H_2}=\dfrac{1}{2}\cdot n_{HCl}=\dfrac{1}{2}\cdot0.5=0.25\left(mol\right)\)
\(m_{Mg}=0.25\cdot24=6\left(g\right)\)
\(V_{H_2}=0.25\cdot22.4=5.6\left(l\right)\)
\(c.\)
\(V_{H_2\left(tt\right)}=5.6\cdot90\%=5.04\left(l\right)\)
a) Fe + 2HCl --> FeCl2 + H2
b) \(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
_____0,4--->0,8------>0,4--->0,4
=> VH2 = 0,4.22,4 = 8,96(l)
c) mHCl = 0,8.36,5 = 29,2 (g)
=> \(m_{dd\left(HCl\right)}=\dfrac{29,2.100}{7,3}=400\left(g\right)\)
mdd (sau pư) = 22,4 + 400 - 0,4.2 = 421,6 (g)
=> \(C\%\left(FeCl_2\right)=\dfrac{127.0,4}{421,6}.100\%=12,05\%\)
Ta có: \(n_{Na_2SO_3}=\dfrac{12,6}{126}=0,1\left(mol\right)\)
a. PTHH: Na2SO3 + 2HCl ---> 2NaCl + SO2 + H2O
Theo PT: \(n_{SO_2}=n_{Na_2CO_3}=0,1\left(mol\right)\)
=> \(V_{SO_2}=0,1.22,4=2,24\left(lít\right)\)
b. Theo PT: \(n_{HCl}=2.n_{SO_2}=2.0,1=0,2\left(mol\right)\)
=> \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
=> \(C_{\%_{HCl}}=\dfrac{7,3}{150}.100\%=4,87\%\)
c. Ta có: \(m_{dd_{NaCl}}=n_{Na_2SO_{3_{PỨ}}}=50\left(g\right)\)
Theo PT: \(n_{NaCl}=n_{HCl}=0,2\left(mol\right)\)
=> \(m_{NaCl}=0,2.58,5=11,7\left(g\right)\)
=> \(C_{\%_{NaCl}}=\dfrac{11,7}{50}.100\%=23,4\%\)
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
a+b+c) Ta có: \(n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)
\(\Rightarrow n_{H_2}=0,25\left(mol\right)=n_{MgCl_2}\) \(\Rightarrow\left\{{}\begin{matrix}m_{MgCl_2}=0,25\cdot95=23,75\left(g\right)\\V_{H_2}=0,25\cdot22,4=5,6\left(l\right)\end{matrix}\right.\)
d) PTHH: \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)
Theo PTHH: \(n_{Cu}=n_{H_2}=0,25\left(mol\right)\) \(\Rightarrow m_{Cu}=0,25\cdot64=16\left(g\right)\)