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nCa(OH)2 = 0,15(mol)
nSO2=0,2(mol)
Ta có: 1< nCa(OH)2/nSO2<2
=> Sp thu được hh 2 muối CaSO3 và Ca(HSO3)2
PTHH: Ca(OH)2 + SO2 -> CaSO3 + H2O (1)
CaSO3 + SO2 + H2O -> Ca(HSO3)2 (2)
Ta có: nSO2(2)= 0,2-0,15=0,05(mol)
=> nCaSO3(2)=0,05(mol)
nCaSO3(1)=nCa(OH)2=0,15(mol)
=>m(kết tủa)= mCaSO3(còn)= (0,15-0,05).120=12(g)
=> Chọn B
\(Fe+2HCl\rightarrow FeCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Đặt:\left\{{}\begin{matrix}n_{Fe}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}56x+27y=2,78\\x+1,5y=0,07\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,04\\y=0,02\end{matrix}\right.\\ n_{FeCl_2}=n_{Fe}=0,04\left(mol\right)\\ \Rightarrow m_{FeCl_2}=0,04.127=5,08\left(g\right)\)
\(n_{H_2}=\frac{5,6}{22,4}=0,25(mol)\\ m_{H_2}=0,25.2=0,5(g)\\ BT H:\\ n_{HCl}=2n_{H_2}=0,25.2=0,5(mol)\\ m_{HCl}=0,5.36,5=18,25(g)\\ BTKL:\\ m_{hh}+m_{HCl}=m_{muối}+m_{H_2}\\ 15+18,25=m_{muối}+0,5\\ \to m_{muối}=32,75(g)\\ \to D\)
\(n_{CH_4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
0,5--------------->0,5
CO2 + Ca(OH)2 --> CaCO3 + H2O
0,5------------------->0,5
=> mCaCO3 = 0,5.100 = 50 (g)
=> B
1.
\(n_{CO_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
\(n_{Ca\left(OH\right)_2}=0.075\left(mol\right)\)
\(T=\dfrac{0.1}{0.075}=1.33\)
=> Tạo ra 2 muối
\(n_{CaCO_3}=a\left(mol\right),n_{Ca\left(HCO_3\right)_2}=b\left(mol\right)\)
Khi đó :
\(a+b=0.075\)
\(a+2b=0.1\)
\(\Rightarrow\left\{{}\begin{matrix}a=0.05\\b=0.025\end{matrix}\right.\)
\(m_{sp}=0.05\cdot100+0.025\cdot162=9.05\left(g\right)\)
2.
\(n_{CO_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)
\(n_{Ba\left(OH\right)_2}=0.2\cdot0.2=0.04\left(mol\right)\)
\(T=\dfrac{0.005}{0.04}=1.25\)
=> Tạo ra 2 muối
\(n_{BaCO_3}=a\left(mol\right),n_{Ba\left(HCO_3\right)_2}=b\left(mol\right)\)
Ta có :
\(a+b=0.04\)
\(a+2b=0.05\)
\(\Rightarrow\left\{{}\begin{matrix}a=0.03\\b=0.01\end{matrix}\right.\)
\(m_{BaCO_3}=0.03\cdot197=5.91\left(g\right)\)
\(n_{CO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
=> nC = 0,4 (mol)
\(n_{H_2O}=\dfrac{9}{18}=0,5\left(mol\right)\)
=> nH = 1 (mol)
m = mC + mH = 0,4.12 + 1.1 = 5,8 (g)
=> A
\(n_{CO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\\ n_{CaCO_3}=n_{CO_2}=0,05\left(mol\right)\\ m_{rắn}=m_{CaCO_3}=0,05.100=5\left(g\right)\\ \Rightarrow ChọnD\)
Ta có: \(n_{CO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
Bảo toàn Cacbon: \(n_{CO_2}=n_{CaCO_3}=0,05\left(mol\right)\) \(\Rightarrow m_{CaCO_3}=0,05\cdot100=5\left(g\right)\)
\(\Rightarrow\) Chọn D