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a.
\(m_{Ag}=m_{k.tan}=8,7\left(g\right)\\ m_{Zn,Mg}=20-8,7=11,3\left(g\right)\\ \left\{{}\begin{matrix}n_{Zn}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\\Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ \rightarrow\left\{{}\begin{matrix}65a+24b=11,3\\a+b=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%m_{Ag}=\dfrac{8,7}{20}.100=43,5\%\\\%m_{Mg}=\dfrac{24.0,2}{20}.100=24\%\\\%m_{Zn}=\dfrac{0,1.65}{20}.100=32,5\%\end{matrix}\right.\)
b.
\(n_{H_2SO_4\left(tổng\right)}=a+b=0,3\left(mol\right)\\ V_{ddH_2SO_4\left(tổng\right)}=\dfrac{0,3}{0,5}=0,6\left(lít\right)=600\left(ml\right)\)
\(n_{H2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
a) Pt : \(Mg+H_2SO_4\rightarrow MgSO_4+H_2|\)
1 1 1 1
0,3 0,3 0,3
b) \(n_{Mg}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
⇒ \(m_{Mg}=0,3.24=7,2\left(g\right)\)
\(n_{H2SO4}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
⇒ \(m_{H2SO4}=0,3.98=29,4\left(g\right)\)
c) \(n_{MgSO4}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
⇒ \(m_{MgSO4}=0,3.120=36\left(g\right)\)
Chúc bạn học tốt
n hh khí = 0.5 mol
nCO: x mol
nCO2: y mol
=> x + y = 0.5
28x + 44y = 17.2 g
=> x = 0.3 mol
y = 0.2 mol
Khối lượng oxi tham gia pứ oxh khử oxit KL: 0.2 * 16 = 3.2g => m KL = 11.6 - 3.2 = 8.4g
TH: KL hóa trị I => nKL = 2*nH2 = 0.3 mol => KL: 28!!
KL hóa trị III => nKL = 2/3 *nH2 = 0.1 mol => KL: 84!!
KL hóa trị II => nKL = nH2 = 0.15 mol => KL: 56 => Fe.
nFe / Oxit = 0.15 mol
nO/Oxit = 0.2 mol
=> nFe/nO = 3/4 => Fe3O4
Fe3O4 + 4CO = 3Fe + 4CO2
Fe + H2SO4 = FeSO4 + H2
0.15.....0.15.......0.15.....0.15
=> mH2SO4 pứ = 14.7 g => mdd = 147 g
m dd sau khi cho KL vào = m KL + m dd - mH2 thoát ra = 0.15 * 56 + 147 - 0.15*2 = 155.1g
=> C% FeSO4 = 14.7%
mFe3O4 = 34.8 / 232 = 0.15 (mol)
nH2 = 8.96 / 22.4 = 0.4 (mol)
Fe3O4 + 4H2 -t0-> 3Fe + 4H2O
Bđ: 0.15.......0.4
Pư: 0.1..........0.4........0.3........0.4
Kt: 0.05.........0............0.3.......0.4
mFe3O4(dư) = 0.05 * 232 = 11.6 (g)
mFe = 0.3 * 56 = 16.8 (g)
VH2O = 0.4 * 22.4 = 8.96 (l)
2Fe + 6H2SO4(đ) => Fe2(SO4)3 + 3SO2 + 6H2O
0.3...........0.9.................0.15...........0.45
mH2SO4 = 0.9 * 98 = 88.2 (g)
C% H2SO4 = 88.2 * 100 / 98 = 90 %
VSO2 = 0.45 * 22.4 = 10.08 (l)
mX = 16.8 + 98 - 0.45 * 64 = 86 (g)
C% Fe2(SO4)3 = 0.15 * 400 / 86 * 100% = 69.76%
Câu 13:
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ PTHH:R_2O_3+3H_2\underrightarrow{t^o}2R+3H_2O\\ Theo.pt:n_{R_2O_3}=\dfrac{1}{3}n_{H_2}=\dfrac{1}{3}.0,3=0,1\left(mol\right)\\ M_{R_2O_3}=\dfrac{16}{0,1}=160\left(\dfrac{g}{mol}\right)\\ \Leftrightarrow2R+16.3=160\\ \Leftrightarrow R=56\left(\dfrac{g}{mol}\right)\\ \Leftrightarrow R.là.Fe\\ CTHH:Fe_2O_3\)
Bài 14:
\(n_{H_2}=\dfrac{2,8}{22,4}=0,125\left(mol\right)\\ PTHH:Fe+H_2SO_{4\left(loãng\right)}\rightarrow FeSO_4+H_2\uparrow\left(1\right)\\ Theo.pt\left(1\right):n_{Fe}=n_{H_2}=0,125\left(mol\right)\\ PTHH:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\left(2\right)\\ Theo.pt\left(2\right):n_{Fe_2O_3}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{3}.0,125=\dfrac{1}{24}\left(mol\right)\\ m=m_{Fe_2O_3}=\dfrac{1}{24}.160=\dfrac{20}{3}\left(g\right)\\ n=n_{Fe}=0,125.56=7\left(g\right)\)
a. PTHH:
Al+H2SO4-->AlSO4+H2
b.Theo ĐLBTKL, ta có:
mAl+mH2SO4=mAl2SO4+mH2
=>mH2SO4=mAl2SO4+mH2-mAl
=171+3-27=147 (g)
a) Ta có : \(m_{KL}+m_{SO^{2-}_4}=m_{muối}\)
=> \(m_{SO_4^{2-}}=8,25-2,49=5,76\left(g\right)\)
=> \(n_{SO_4^{2-}}=\dfrac{5,76}{96}=0,06\left(mol\right)\)
Mặc khác : \(2H^++SO_4^{2-}\rightarrow H_2SO_4\)
=>\(n_{SO_4^{2-}}=n_{H_2SO_4}=0,06\left(mol\right)\)
=> \(m_{H_2SO_4}=0,06.98=5,88\left(g\right)\)
b) Bảo toàn nguyên tố H : \(n_{H_2}=n_{H_2SO_4}=0,06\left(mol\right)\)
=> VH2 = 0,06.22,4 = 1,344(lít )
\(\left\{{}\begin{matrix}Zn\\Fe\\Mg\end{matrix}\right.+H_2SO_4\rightarrow\left\{{}\begin{matrix}ZnSO_4\\FeSO_4\\MgSO_4\end{matrix}\right.+H_2\uparrow\)
Ta có: \(m_{SO_4}=8,25-2,49=5,76\left(g\right)\)
\(\Rightarrow n_{H_2SO_4}=n_{H_2}=n_{SO_4}=\dfrac{5,76}{96}=0,06\left(mol\right)\)
a, \(m_{H_2SO_4}=0,06.98=5,88\left(g\right)\)
b, \(V_{H_2}=0,06.22,4=1,344\)
\(a.n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\\ n_{Mg}=a,n_{Na}=b\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ 2Na+2H_2O\rightarrow2NaOH+H_2\\ 2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\\ \Rightarrow\left\{{}\begin{matrix}24a+23b=9,4\\a+0,5b=0,3\end{matrix}\right.\\ \Rightarrow a=b=0,2\\ m_{Mg}=0,2.24=4,8g\\ m_{Na}=0,2.23=3,6g\\ b.n_{H_2SO_4}=0,5b+a=0,3mol\\ m_{ddH_2SO_4}=\dfrac{0,3.98}{10\%}\cdot100\%=294g\\ c.n_{MgSO_4}=n_{Mg}=0,2mol\\ n_{Na_2SO_4}=0,5n_{Na}=0,1mol\\ m_{MgSO_4}=120.0,2=24g\\ m_{Na_2SO_4}=142.0,1=14,2g\)