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a) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
____0,15<--0,3<-------------0,15
=> mFe = 0,15.56 = 8,4 (g)
b) \(C_{M\left(ddHCl\right)}=\dfrac{0,3}{0,05}=6M\)
Bảo toàn Hidro: \(n_{HCl}=2n_{H_2}=2\cdot\dfrac{3,36}{22,4}=0,3\left(mol\right)\) \(\Rightarrow C_{M_{HCl}}=\dfrac{0,3}{0,5}=0,6\left(M\right)\)
\(\Rightarrow\) Chọn A
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
\(n_{Fe}=n_{H_2}=0,15\left(mol\right)\)
\(m_{Fe}=0,15.56=8,4\left(g\right)\)
\(n_{HCl}=2n_{H_2}=0,3\left(mol\right)\)
\(50ml=0,05l\)
\(C_{M_{ddHCl}}=\dfrac{0,3}{0,05}=6\left(M\right)\)
\(n_{H2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
a) Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)
1 2 1 1
0,15 0,3 0,15
b) \(n_{Fe}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)
⇒ \(m_{Fe}=0,15.56=8,4\left(g\right)\)
c) \(n_{HCl}=\dfrac{0,15.2}{1}=0,3\left(mol\right)\)
50ml = 0,05l
\(C_{M_{ddHCl}}=\dfrac{0,3}{0,05}=6\left(M\right)\)
Chúc bạn học tốt
nH2=3.36/22.4=0.15 mol
a) PT: Fe + 2HCl ----> FeCl2 + H2
0.15 0.3 0.15
b)mFe=0.15*56=8.4g
c)CMHCl= 0.3*0.05=6 M
Chúc em học tốt!!!
Fe+2HCl->FeCl2+H2
nH2=0.15(mol)
Theo pthh nFe=nH2->nFe=0.15(mol)
mFe phản ứng:0.15*56=8.4(g)
nHCl=2nH2->nHCl=0.3(mol)
CM=0.3:0.05=6 M
a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Theo PT: \(n_{Fe}=n_{FeCl_2}=n_{H_2}=0,3\left(mol\right)\Rightarrow m_{Fe}=0,3.56=16,8\left(g\right)\)
b, \(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,15}=4\left(M\right)\)
c, \(FeCl_2+2NaOH\rightarrow2NaCl+Fe\left(OH\right)_2\)
Theo PT: \(n_{NaOH}=2n_{FeCl_2}=0,6\left(mol\right)\)
\(\Rightarrow V_{NaOH}=\dfrac{0,6}{1}=0,6\left(l\right)=600\left(ml\right)\)
\(a.Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe,pư}=n_{FeCl_2}=n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\\ m_{Fe,pư}=0,3.56=16,8g\\ b.n_{HCl}=0,3.2=0,6mol\\ C_{M_{HCl}}=\dfrac{0,6}{0,15}=4M\\ c.2NaOH+FeCl_2\rightarrow Fe\left(OH\right)_2+2NaCl\\ n_{NaOH}=0,3.2=0,6mol\\ V_{ddNaOH}=\dfrac{0,6}{1}=0,6l=600ml\)
\(n_{H2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)
1 2 1 1
0,15 0,3 0,15
\(n_{Fe}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)
⇒ \(m_{Fe}=0,15.56=8,4\left(g\right)\)
\(n_{HCl}=\dfrac{0,15.2}{1}=0,3\left(mol\right)\)
50ml = 0,05l
\(C_{M_{HCl}}=\dfrac{0,3}{0,05}=6\left(M\right)\)
Chúc bạn học tốt
Chọn B