Câu 2: Cho 5,6 gam Fe phản ứng với Clo vừa đủ sinh ra m gam muối.
a) Viết phương trình phản ứng
b) Tính thể tích Clo đã phản ứng.
c) Tính khối lượng muối tạo thành.
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a) $2Al + 3Cl_2 \xrightarrow{t^o} 2AlCl_3$
b) $n_{Al} = \dfrac{5,4}{27} = 0,2(mol)$
Theo PTHH : $n_{Cl_2} = \dfrac{3}{2}n_{Al} = 0,3(mol)$
$\Rightarrow V_{Cl_2} = 0,3.24,79 = 7,437(lít)$
c) $n_{AlCl_3} = n_{Al} = 0,2(mol)$
$\Rightarrow m_{AlCl_3} = 0,2.133,5 = 26,7(gam)$
Ta có: \(n_{NaCl}=\dfrac{5,85}{58,5}=0,1\left(mol\right)\)
PT: \(2Na+Cl_2\rightarrow2NaCl\)
____0,1___0,05____0,1 (mol)
a, mNa = 0,1.23 = 2,3 (g)
b, VCl2 = 0,05.22,4 = 1,12 (l)
Bạn tham khảo nhé!
Câu 1:
\(Mg+Br_2\rightarrow MgBr_2\\ n_{Br_2}=\dfrac{11,2}{160}=0,07\left(mol\right)=n_{Mg}=n_{MgBr_2}\\ a=m_{Mg}=0,07.24=1,68\left(g\right)\\ m_{MgBr_2}=184.0,07=12,88\left(g\right)\)
\(a.Sắt+Clo\rightarrow Sắt\left(III\right)clorua\\ b.2Fe+3Cl_2-^{t^o}\rightarrow2FeCl_3\\ c.m_{Fe}+m_{Cl_2}=m_{FeCl_3}\\ \Rightarrow m_{FeCl_3}=5,6+10,65=16,25\left(g\right)\)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right);n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\a, Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,V\text{ì}:\dfrac{0,2}{1}>\dfrac{0,1}{1}\Rightarrow Zn\text{dư}\\ \Rightarrow n_{Zn\left(p.\text{ứ}\right)}=n_{ZnCl_2}=n_{H_2}=0,1\left(mol\right)\\b, m_{Zn\left(p.\text{ứ}\right)}=0,1.65=6,5\left(g\right)\\ n_{HCl}=0,1.2=0,2\left(mol\right)\\ m_{HCl}=0,2.36,5=7,3\left(g\right)\\ d,m_{ZnCl_2}=136.0,1=13,6\left(g\right)\)
Chúc em học tốt !!!
\(a.\)
\(n_{Na}=\dfrac{4.6}{23}=0.2\left(mol\right)\)
\(Na+\dfrac{1}{2}Cl_2\underrightarrow{t^0}NaCl\)
\(0.2........0.1........0.2\)
\(V_{Cl_2}=0.1\cdot22.4=2.24\left(l\right)\)
\(m_{NaCl}=0.2\cdot58.5=11.7\left(g\right)\)
\(b.\)
\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)
\(Fe+\dfrac{3}{2}Cl_2\underrightarrow{t^0}FeCl_3\)
\(0.1.......0.15.......0.1\)
\(V_{Cl_2}=0.15\cdot22.4=3.36\left(l\right)\)
\(m_{FeCl_3}=0.1\cdot162.5=16.25\left(g\right)\)
\(c.\)
\(n_{Cu}=\dfrac{6.4}{64}=0.1\left(mol\right)\)
\(Cu+Cl_2\underrightarrow{t^0}CuCl_2\)
\(0.1......0.1.....0.1\)
\(V_{Cl_2}=0.1\cdot22.4=2.24\left(l\right)\)
\(m_{CuCl_2}=0.1\cdot135=13.5\left(g\right)\)
Bài 1:
a. \(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
\(2Na+Cl_2\rightarrow2NaCl\)
0,2 ...... 0,1 ..... 0,2 (mol)
\(\rightarrow\left\{{}\begin{matrix}V_{Cl_2}=0,1.22,4=2,24\left(l\right)\\m_{NaCl}=0,2.58,5=11,7\left(g\right)\end{matrix}\right.\)
b. \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(2Fe+3Cl_2\rightarrow2FeCl_3\)
0,1 ...... 0,15 ...... 0,1 (mol)
\(\rightarrow\left\{{}\begin{matrix}V_{Cl_2}=0,15.22,4=3,36\left(l\right)\\m_{FeCl_3}=0,1.162,5=16,25\left(g\right)\end{matrix}\right.\)
c. \(n_{Cu}=\dfrac{6,4}{64}=0,1\left(mol\right)\)
\(Cu+Cl_2\rightarrow CuCl_2\)
0,1 .... 0,1 ..... 0,1 (mol)
\(\rightarrow\left\{{}\begin{matrix}V_{Cl_2}=0,1.22,4=2,24\left(l\right)\\m_{CuCl_2}=0,1.135=13,5\left(g\right)\end{matrix}\right.\)
\(a.Fe+2HCl\rightarrow FeCl_2+H_2\\b.n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ n_{H_2}=n_{Fe}=0,1\left(mol\right)\\ \Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\\ c.n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\\ m_{FeCl_2}=0,1.127=12,7\left(g\right) \)
2Al+3H2SO4->al2(SO4)3+3H2
Fe+H2SO4->FeSO4+H2
Gọi x,y tương ứng là số mol của Al và Fe:
Ta có: 27x+56y=11 (1)
nH2=0,4 mol
1,5x+y=0,4 (2)
Giải hệ(1),(2):x=0,2;y=0,1
mAl=0,2.27=5,4g
%Al=\(\dfrac{5,4.100}{16,6}\)=32,53%
=>%Fe=67,47%
m H2SO4=0,4.98=39,2g
c) m muối=0,1.342+0,1.152=49,4g
Tham khảo:
PTHH: 2Fe + 3Cl2--> 2FeCl3
Ta có: nFe=5,6/56=0,1 mol
Theo PTHH ta có:
nFeCl3 = nFe , nCl2=3nFe/2=0,15 mol
=> VCl2=0,15.22,4=3,36 l
mFeCl3=162,5.0,1=16,25 g
\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)
\(Fe+\dfrac{3}{2}Cl_2\underrightarrow{^{^{t^0}}}FeCl_3\)
\(0.1......0.15........0.1\)
\(V_{Cl_2}=0.15\cdot22.4=3.36\left(l\right)\)
\(m_{FeCl_3}=0.1\cdot162.5=16.25\left(g\right)\)