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a. Đổi 200 ml = 0,2 lít
\(n_{Fe}=\dfrac{11.2}{56}=0,2\left(mol\right)\)
\(n_{HCl}=2.0,2=0,2\left(mol\right)\)
PTHH : Fe + 2HCl -> FeCl2 + H2
0,1 0,2 0,1 0,1
Ta thấy : \(\dfrac{0.2}{1}>\dfrac{0.2}{2}\) => Fe dư , HCl đủ
\(m_{Fe\left(dư\right)}=\left(0,2-0,1\right).56=5,6\left(g\right)\)
b. \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c. Sau phản ứng chất tan là FeCl2
\(V_{FeCl_2}=0,1.2=0,2\left(l\right)\)
\(\Rightarrow C_{M_{FeCl_2}}=\dfrac{0.1}{0,2}=0,5\left(M\right)\)
a, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
\(n_{HCl}=0,1.2=0,2\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,2}{2}\), ta được Fe dư.
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
b, Fe dư.
Theo PT: \(n_{Fe\left(pư\right)}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\Rightarrow n_{Fe\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\)
\(\Rightarrow m_{Fe\left(dư\right)}=0,1.56=5,6\left(g\right)\)
c, \(n_{FeCl_2}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\Rightarrow C_{M_{FeCl_2}}=\dfrac{0,1}{0,1}=1\left(M\right)\)
\(a,n_{HCl}=0,1.1=0,1\left(mol\right)\\ n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PTHH: Fe + 2HCl ---> FeCl2 + H2
LTL: \(0,1>\dfrac{0,1}{2}\) => Fe dư
Theo pthh: \(n_{H_2}=n_{FeCl_2}=n_{Fe\left(pư\right)}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)
=> VH2 = 0,05.22,4 = 1,12 (l)
b, Chất dư là Fe
mFe (dư) = (0,1 - 0,05).56 = 2,8 (g)
c, \(C_{M\left(FeCl_2\right)}=\dfrac{0,05}{0,1}=0,5M\)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ n_{H_2SO_4}=0,2.2=0,4\left(mol\right)\\ Vì:\dfrac{0,2}{1}< \dfrac{0,4}{1}\Rightarrow H_2SO_4dư\\ n_{H_2}=n_{H_2SO_4\left(p.ứ\right)}=n_{FeSO_4}=n_{Fe}=0,2\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ b,n_{H_2SO_4\left(dư\right)}=0,4-0,2=0,2\left(mol\right)\\ m_{H_2SO_4}=0,2.98=19,6\left(g\right)\\ c,V_{ddsau}=V_{ddH_2SO_4}=200\left(ml\right)=0,2\left(l\right)\\ C_{MddH_2SO_4\left(dư\right)}=\dfrac{0,2}{0,2}=1\left(M\right)\\ C_{MddFeSO_4}=\dfrac{0,2}{0,2}=1\left(M\right)\)
\(n_{CaCO_3}=\dfrac{10}{100}=0.1\left(mol\right)\)
\(n_{HCl}=\dfrac{3.65}{36.5}=0.1\left(mol\right)\)
\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)
\(1..................2\)
\(0.1..............0.1\)
\(LTL:\dfrac{0.1}{1}>\dfrac{0.1}{2}\Rightarrow CaCO_3dư\)
\(m_{CaCO_3\left(dư\right)}=\left(0.1-0.05\right)\cdot100=5\left(g\right)\)
\(V_{CO_2}=0.05\cdot22.4=1.12\left(l\right)\)
Để phản ứng xảy ra vừa đủ :
\(n_{CaCO_3\left(dư\right)}=\dfrac{5}{100}=0.05\left(mol\right)\)
\(n_{HCl}=2\cdot0.05=0.1\left(mol\right)\)
\(m_{HCl\left(ct\right)}=0.1\cdot36.5=3.65\left(g\right)\)
tham khảo ở đây nha bạn:https://hoidap247.com/cau-hoi/981113
a, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
\(n_{H_2SO_4}=0,2.2=0,4\left(mol\right)\)
PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,4}{1}\), ta được H2SO4 dư.
Theo PT: \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b, - H2SO4 dư.
\(n_{H_2SO_4\left(pư\right)}=n_{Fe}=0,2\left(mol\right)\Rightarrow n_{H_2SO_4\left(dư\right)}=0,4-0,2=0,2\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4\left(dư\right)}=0,2.98=19,6\left(g\right)\)
c, \(C_{M_{H_2SO_4\left(dư\right)}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
Theo PT: \(n_{FeSO_4}=n_{Fe}=0,2\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
\(a,n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right);n_{HCl}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\\ CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ Vì:\dfrac{0,1}{1}>\dfrac{0,1}{2}\Rightarrow CaCO_3dư\\ n_{CO_2}=n_{CaCO_3\left(p.ứ\right)}=\dfrac{n_{HCl}}{2}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ \Rightarrow n_{CaCO_3\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\\ \Rightarrow m_{CaCO_3\left(dư\right)}=0,05.100=5\left(g\right)\\ b,V_{CO_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\)
c) Muốn phản ứng xảy ra vừa đủ, chất nào thiếu ta cần thêm chất đó vào => Cần thêm HCl vào.
\(n_{HCl\left(cần\right)}=2.n_{CaCO_3}=2.0,1=0,2\left(mol\right)\\ \Rightarrow n_{HCl\left(thêm\right)}=0,2-0,1=0,1\left(mol\right)\\ \Rightarrow m_{HCl\left(thêm\right)}=0,1.36,5=3,65\left(g\right)\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right);n_{HCl}=1.0,1=0,1\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ Vì:\dfrac{0,1}{2}< \dfrac{0,1}{1}\Rightarrow Fe.dư\\ n_{H_2}=n_{FeCl_2}=n_{Fe\left(p.ứ\right)}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\\ b,n_{Fe\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\\ m_{Fe\left(dư\right)}=0,05.56=2,8\left(g\right)\\ c,V_{ddFeCl_2}=V_{ddHCl}=0,1\left(l\right)\\ C_{MddFeCl_2}=\dfrac{0,05}{0,1}=0,5\left(M\right)\)
1/ nFe= 11.2/56=0.2 mol
nHCl= 0.1*2=0.2 mol
Fe + 2HCl --> FeCl2 + H2
Bđ: 0.2___0.2
Pư: 0.1___0.2____0.1____0.1
Kt: 0.1____0_____0.1____0.1
VH2= 0.1*22.4=2.24l
mFe (dư)= 0.1*56=5.6g
CM FeCl2= 0.1/0.1=1M
2/
nCaCO3= 10/100=0.1 mol
nHCl= 5.475/36.5=0.15 mol
CaCO3 + 2HCl --> CaCl2 + CO2 + H2O
Bd: 0.1 _____0.15
Pư: 0.075___0.15____________0.15
Kt: 0.025____0______________0.15
mCaCO3 dư= 0.025*100=2.5g
VCO2= 0.15*22.4=3.36l
Muốn phản ứng vừa đủ phải thêm vào dd HCl
CaCO3 + 2HCl --> CaCl2 + CO2 + H2O
0.025____0.05
mHCl cần thêm= 0.05*36.5=1.825g
các từ bđ kt là j??