Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(2Fe+6H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)
Gọi 0,375a là mol Fe \(\Rightarrow\) nH2SO4= a mol
\(n_{Fe_{pu}}=\frac{a}{3}\left(mol\right)\)
Nên dư a/24 mol Fe. Tạo a/6 mol Fe2(SO4)3
\(Fe+Fe_2\left(SO_4\right)_3\rightarrow3FeSO_4\)
\(\Rightarrow\) nFeSO4= a/8 mol. Dư a/8 mol Fe2(SO4)3.
m muối= 8,28g
\(\Rightarrow\frac{152a}{8}+\frac{400a}{8}=8,28\)
\(\Rightarrow a=0,12\)
nFe phản ứng= 0,375a= 0,045 mol
\(\Rightarrow m_{Fe}=2,52\left(g\right)\)
\(n_{H2SO4}=0,12\left(mol\right)\Rightarrow n_{SO2}=0,06\left(mol\right)\)
\(n_{NaOH}=0,1\left(mol\right)\)
\(\frac{n_{NaOH}}{n_{SO2}}=1,67\Rightarrow\) Tạo 2 muối
\(NaOH+SO_2\rightarrow NaHSO_3\)
\(2NaOH+SO_2\rightarrow Na_2SO_3+H_2O\)
Gọi x là mol NaHSO3, y là mol Na2SO3
\(\left\{{}\begin{matrix}x+2y=0,1\\x+y=0,06\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,02\\y=0,04\end{matrix}\right.\)
\(CM_{NaHSO3}=\frac{0,02}{0,1}=0,2M\)
\(CM_{Na2SO3}=\frac{0,04}{0,1}=0,4M\)
\(a.n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,2 0,3 0,1 0,3
\(m_{Al}=0,2.27=5,4g\\ b.C_{M\left(H_2SO_4\right)}=\dfrac{0,3}{0,45}=\dfrac{2}{3}M\\ c.2H_2+O_2\underrightarrow{t^0}2H_2O\)
0,3 0,15 0,3
\(V_{O_2}=0,15.22,4=3,36l\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}.0,3=0,2\left(mol\right)\\ a,m_{Al}=0,2.27=5,4\left(g\right)\\ n_{H_2SO_4}=n_{H_2}=0,3\left(mol\right)\\ b,C_{MddH_2SO_4}=\dfrac{0,3}{0,45}=\dfrac{2}{3}\left(M\right)\\ 2H_2+O_2\rightarrow\left(t^o\right)2H_2O\\ n_{O_2}=\dfrac{n_{H_2}}{2}=\dfrac{0,3}{2}=0,15\left(mol\right)\\ c,V_{O_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\)
\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{3,36}{22,4}=0,15mol\)
a. PTHH: Fe + H2SO4 \(\rightarrow\) FeSO4 + H2
TL: 1 1 1 1
mol: 0,15 \(\leftarrow\) 0,15 \(\leftarrow\) 0,15 \(\leftarrow\) 0,15
\(b.m_{Fe}=n.M=0,15.56=8,4g\)
Đổi 150ml = 0,15 l
\(c.C_{MddH_2SO_4}=\dfrac{n}{V}=\dfrac{0,15}{0,15}=1M\)
\(a,n_{H_2SO_4}=0,5\cdot0,1=0,05\left(mol\right)\\ PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ \Rightarrow n_{H_2}=n_{H_2SO_4}=0,05\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,05\cdot22,4=1,12\left(l\right)\\ b,n_{Al}=\dfrac{2}{3}n_{H_2SO_4}=\dfrac{1}{30}\left(mol\right)\\ \Rightarrow m_{Al}=\dfrac{1}{30}\cdot27=0,9\left(g\right)\\ c,n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2SO_4}\approx0,017\left(mol\right)\\ \Rightarrow C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0,017}{0,1}\approx0,17M\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
Ta có: \(n_{H_2}=\dfrac{3,7185}{22,4}\approx0,166\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=0,166\left(mol\right)\\n_{HCl}=0,332\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,166\cdot56=9,296\left(g\right)\\C_{M_{HCl}}=\dfrac{0,332}{0,15}\approx2,21\left(M\right)\end{matrix}\right.\)
a, \(n_{H_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)
PTHH: Fe + 2HCl → FeCl2 + H2
Mol: 0,15 0,3 0,15
\(m_{Fe}=0,15.56=8,4\left(g\right)\)
b, \(C_{M_{ddHCl}}=\dfrac{0,3}{0,15}=2M\)
2Al+ 3H2SO4→ Al2(SO4)3+ 3H2
(mol) 0,1 0,15 0,05 0,15
đổi: 300ml=0,3 lít
a) nAl=\(\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(n_{H_2SO_4}=C_M.V=0,5.0,3=0,15\left(mol\right)\)
tỉ lệ:
Al H2SO4
\(\dfrac{0,2}{2}\) > \(\dfrac{0,15}{3}\)
→ Al dư, H2SO4 phản ứng hết sau phản ứng
→ \(V_{H_2}=n.22,4=0,15.22,4=3,36\left(lít\right)\)
b) \(n_{Al\left(dư\right)}=n_{Al\left(bđ\right)}-n_{Al\left(ph.ứ\right)}=0,2-0,1=0,1\left(mol\right)\)
\(C_{M_{Al\left(dư\right)}}=\dfrac{n}{V}=\dfrac{0,1}{0,3}=\dfrac{1}{3}M\)
\(C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{n}{V}=\dfrac{0,05}{0,3}=\dfrac{1}{6}M\)
\(C_{M_{H_2}}=\dfrac{n}{V}=\dfrac{0,15}{0,3}=0,05M\)
a.
2Fe + 6H2SO4 —> Fe2(SO4)3 + 3SO2 + 6H2O
2x 6x x 3x
Fe + Fe2(SO4)3 —> 3FeSO4
y y 3y
\(\rightarrow\) nFe = 2x + y = 37,5%.6x
m muối = 400(x - y) + 152 . 3y = 8,28
\(\rightarrow\) x = 0,02 và y = 0,005
\(\rightarrow\) nFe = 2x + y = 0,045
\(\rightarrow\)mFe = 2,52
b.
nNaOH = CM . V = 1 . 0,1 = 0,1 mol
nSO2 = 3x = 0,06mol
2NaOH + SO2 ---> Na2SO3 + H2O
0,1 0,05 0,05
Na2SO3 + SO2 + H2O ----> 2NaHSO3
0,01 0,01 0,02
\(\rightarrow\)dd sau phản ứng gồm: Na2SO3: 0,04mol và NaHSO3: 0,02mol
CM Na2SO3 = 0,04 : 0,1 = 0,4M
CM NaHSO3 = 0,02 : 0,1 = 0,2M
Tham khảo