Hòa tan 9 gam hợp kim nhôm- magie vào dung dịch H2SO4 dư thu được 10,08 lít khí H2 (đktc). Tính thành phần % khối lượng của nhôm và magie trong hợp kim.
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Câu 1:
Ta có: 1 tấn = 1000 kg
⇒ mFe = 1000.95% = 950 (kg)
\(\Rightarrow n_{Fe}=\dfrac{950}{56}\left(kmol\right)\)
BTNT Fe, có: \(n_{Fe_2O_3\left(LT\right)}=\dfrac{1}{2}n_{Fe}=\dfrac{475}{56}\left(kmol\right)\)
Mà: H = 80% \(\Rightarrow n_{Fe_2O_3\left(TT\right)}=\dfrac{475}{56}:80\%=\dfrac{2375}{224}\left(kmol\right)\)
\(\Rightarrow m_{Fe_2O_3\left(TT\right)}=\dfrac{2375}{224}.160=\dfrac{11875}{7}\left(kg\right)\)
⇒ m quặng \(=\dfrac{m_{Fe_2O_3}}{60\%}\approx2827,38\left(kg\right)\)
Câu 2:
Ta có: 65nZn + 27nAl = 3,79 (1)
PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
Theo PT: \(n_{H_2}=n_{Zn}+\dfrac{3}{2}n_{Al}=\dfrac{1,792}{22,4}=0,08\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Zn}=0,05\left(mol\right)\\n_{Al}=0,02\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Zn}=0,05.65=3,25\left(g\right)\\m_{Al}=0,02.27=0,54\left(g\right)\end{matrix}\right.\)
Câu 21:
Ta có: m1 + m2 = 60 (1)
\(m_{NaOH\left(10\%\right)}=10\%.m_1\left(g\right)\)
\(m_{NaOH\left(40\%\right)}=40\%.m_2\left(g\right)\)
\(\Rightarrow10\%m_1+40\%m_2=60.20\%\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}m_1=40\left(g\right)\\m_2=20\left(g\right)\end{matrix}\right.\)
→ Đáp án: C
Câu 22:
Gọi: \(\left\{{}\begin{matrix}m_{CuSO_4.5H_2O}=a\left(g\right)\\m_{ddCuSO_4\left(8\%\right)}=b\left(g\right)\end{matrix}\right.\) ⇒ a + b = 280 (1)
Ta có: \(n_{CuSO_4}=n_{CuSO_4.5H_2O}=\dfrac{a}{250}\left(mol\Rightarrow\right)m_{CuSO_4}=\dfrac{a}{250}.160=\dfrac{16a}{25}\left(g\right)\)
\(m_{CuSO_4\left(8\%\right)}=8\%b\left(g\right)\)
\(\Rightarrow\dfrac{16a}{25}+8\%b=280.16\%\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=40\left(g\right)\\b=240\left(g\right)\end{matrix}\right.\)
→ Đáp án: D
Câu 23:
Ta có: \(n_{HCl\left(4M\right)}=\dfrac{V}{1000}.4\left(mol\right)\)
\(n_{HCl\left(0,5M\right)}=0,4.0,5=0,2\left(mol\right)\)
\(\Rightarrow\dfrac{V}{1000}.4+0,2=\dfrac{V+400}{1000}.2\)
\(\Rightarrow V=300\left(ml\right)\)
→ Đáp án: C
Lần sau bạn đăng tách từng câu ra nhé.
Câu 16:
Ta có: \(d_{hh/H_2}=14,5\Rightarrow\overline{M}_{hh}=4,5.2=29\left(g/mol\right)\)
\(\Rightarrow\dfrac{28n_{CO}+30n_{NO}}{n_{CO}+n_{NO}}=29\)
\(\Rightarrow n_{CO}=n_{NO}\)
\(\Rightarrow\%V_{CO}=\%V_{NO}=50\%\)
→ Đáp án: A
Câu 17:
Có: \(d_{hh/H_2}=8,375\Rightarrow\overline{M}_{hh}=8,375.4=33,5\left(g/mol\right)\)
\(\Rightarrow\dfrac{44n_{N_2O}+30n_{NO}}{n_{N_2O}+n_{NO}}=33,5\)
\(\Rightarrow n_{N_2O}=\dfrac{1}{3}n_{NO}\)
\(\Rightarrow\%V_{NO}=\dfrac{n_{NO}}{n_{N_2O}+n_{NO}}.100\%=\dfrac{n_{NO}}{\dfrac{1}{3}n_{NO}+n_{NO}}.100\%=75\%\)
→ Đáp án: D
Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Theo PT: \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1.24,79=2,479\left(l\right)\)
→ Đáp án: B
\(n_{Al_2O_3}=\dfrac{2,04}{102}=0,02\left(mol\right)\)
\(n_{HCl}=0,4V_1\left(mol\right)\)
\(n_{NaOH}=1,2V_2\left(mol\right)\)
Ta có: V1 + V2 = 0,5 (1)
TH1: HCl dư.
PT: \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
\(NaOH+HCl\rightarrow NaCl+H_2O\)
Theo PT: nHCl = 6nAl2O3 + nNaOH
⇒ 0,4V1 = 0,02.6 + 1,2V2 (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}V_1=0,45\left(l\right)\\V_2=0,05\left(l\right)\end{matrix}\right.\)
TH2: NaOH dư.
PT: \(Al_2O_3+2NaOH\rightarrow2NaAlO_2+H_2O\)
\(HCl+NaOH\rightarrow NaCl+H_2O\)
Theo PT: nNaOH = nHCl + 2nAl2O3
⇒ 1,2V2 = 0,4V1 + 0,02.2 (3)
Từ (1) và (3) \(\Rightarrow\left\{{}\begin{matrix}V_1=0,35\left(l\right)\\V_2=0,15\left(l\right)\end{matrix}\right.\)
a) \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
nFe = nH2 = 0,3 (mol)
\(\Rightarrow m_{Fe}=0,3.56=16,8\left(g\right)\)
b) nHCl = 2.nH2 = 0,6 (mol)
\(\Rightarrow V_{ddHCl}=\dfrac{0,6}{0,3}=2\left(l\right)\)
c) \(n_{FeCl_2}=n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow C_{M\left(FeCl_2\right)}=\dfrac{0,3}{2}=0,15M\)
Ta có: \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
\(n_{N_2}=\dfrac{1,792}{22,4}=0,08\left(mol\right)\)
BT e, có: 3nAl = 10nN2 + 8nNH4+
⇒ nNH4+ = 0,05 (mol)
BTNT Al, có: nAl(NO3)3 = nAl = 0,4 (mol)
⇒ m muối = 0,4.213 + 0,05.80 = 89,2 (g)
nHNO3 = 12nN2 + 10nNH4+ = 1,46 (mol)
\(\Rightarrow V_{HNO_3}=\dfrac{1,46}{2}=0,73\left(l\right)=730\left(ml\right)\)
Ta có: \(n_{Mg}=\dfrac{21,6}{24}=0,9\left(mol\right)\)
\(n_{N_2O}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
BT e, có: 2nMg = 8nN2O + 8nNH4+
⇒ nNH4+ = 0,075 (mol)
BTNT Mg: nMg(NO3)2 = nMg = 0,9 (mol)
⇒ m = mMg(NO3)2 + mNH4NO3 = 0,9.148 + 0,075.80 = 139,2 (g)
Ta có: nHNO3 = 10nN2O + 10nNH4+ = 0,15.10 + 0,075.10 = 2,25 (mol)
\(\Rightarrow V_{HNO_3}=\dfrac{2,25}{1}=2,25\left(l\right)\)
a, Ta có: \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)
\(n_{N_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)
BT e, có: 3nAl = 10nN2 + 8nNH4+
⇒ nNH4+ = 0,075 (mol)
BTNT Al, có: nAl(NO3)3 = nAl = 0,3 (mol)
⇒ mmuối = mAl(NO3)3 + mNH4NO3 = 0,3.213 + 0,075.80 = 69,9 (g)
b, nHNO3 = 12nN2 + 10nNH4+ = 0,03.12 + 0,075.10 = 1,11 (mol)
Ta có: 27nAl + 24nMg = 9 (1)
PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Mg}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,2\left(mol\right)\\n_{Mg}=0,15\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{9}.100\%=60\%\\\%m_{Mg}=40\%\end{matrix}\right.\)
\(Đặt:n_{Al}=a\left(mol\right);n_{Mg}=b\left(mol\right)\left(a,b>0\right)\\ n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\\ PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ \Rightarrow\left\{{}\begin{matrix}27a+24b=9\\1,5a+b=0,45\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,15\end{matrix}\right.\\ \Rightarrow\%m_{Al}=\dfrac{0,2.27}{9}.100\%=60\%;\%m_{Mg}=\dfrac{24.0,15}{9}.100\%=40\%\)