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nH2SO4=0,4*1=0,4 mol
nH2=6,72/22,4=0,3 mol
2Al + 3H2SO4 -->Al2(So4)3 + 3H2
0,2 0,3 mol
=> mAl = 0,2*27=5,4 g
=> mCu =5,9-5,4=0,5 g
BaCl2 + H2SO4 --> BaSO4 + 2HCl
0,4 0 ,4 mol
=> m BaSO4 = 0,4 * 233=93,2 g
\(a) 2Al+ 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2\\ n_{H_2} = \dfrac{6,72}{22,4} = 0,3(mol)\\ n_{Al} = \dfrac{2}{3}n_{H_2} = 0,2(mol)\\ m_{Al} = 0,2.27 = 5,4(gam)\\ m_{Cu} = 5,9 - 5,4 = 0,5(gam)\\ b) \)
Bảo toàn nguyên tố với S :
\(n_{BaSO_4} = n_{H_2SO_4} = 0,4(mol)\\ m = 0,4.233 = 93,2(gam)\)
a) Đặt \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow27a+24b=1,26\) (1)
Ta có: \(n_{H_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)
Bảo toàn electron: \(3a+2b=0,12\) (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=n_{Al}=0,02\left(mol\right)\\b=n_{Mg}=0,03\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,02\cdot27}{1,26}\cdot100\%\approx42,86\%\\\%m_{Mg}=57,14\%\end{matrix}\right.\)
b) Bảo toàn nguyên tố: \(\left\{{}\begin{matrix}n_{AlCl_3}=n_{Al}=0,02\left(mol\right)\\n_{MgCl_2}=n_{Mg}=0,03\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{AlCl_3}=0,02\cdot133,5=2,67\left(g\right)\\m_{MgCl_2}=0,03\cdot95=2,85\left(g\right)\end{matrix}\right.\)
Mặt khác: \(\left\{{}\begin{matrix}m_{ddHCl}=40\cdot1,25=50\left(g\right)\\m_{H_2}=0,06\cdot2=0,12\left(g\right)\end{matrix}\right.\)
\(\Rightarrow m_{dd}=m_{KL}+m_{ddHCl}-m_{H_2}=51,14\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{AlCl_3}=\dfrac{2,67}{51,14}\cdot100\%\approx5,22\%\\C\%_{MgCl_2}=\dfrac{2,85}{51,14}\cdot100\%\approx5,57\%\end{matrix}\right.\)
\(a)Bte:3n_{Fe}+3n_{Al}=n_{Ag}\\ \Leftrightarrow n_{Ag_3}=0,1.3+0,1.3\\ \Leftrightarrow n_{Ag}=0,6mol\\ m_{rắn}=m_{Ag}=0,6.108=64,8g\\ BTNT\left(Ag\right):n_{Ag}=n_{AgNO_3}=0,6mol\\ V_{AgNO_3}=\dfrac{0,6}{2}=0,3l\\ BTNT\left(Al\right):n_{Al}=2n_{Al_2O_3}\\ \Leftrightarrow0,1=2n_{Al_2O_3}\\ \Leftrightarrow n_{Al_2O_3}=0,05mol\\ BTNT\left(Fe\right):n_{Fe}=2n_{Fe_2O_3}\\ \Leftrightarrow0,1=2n_{Fe_2O_3}\\ \Leftrightarrow n_{Fe_2O_3}=0,05mol \\ b=m_{oxit.bazo}=0,05.\left(160+102\right)=13,1g\)
Gọi số mol NaBr là x; CaCl2 là y \(\rightarrow m=103x+111y\)
\(NaBr+AgNO_3\rightarrow AgBr+NaNO_3\)
\(CaCl_2+2AgNO_3\rightarrow2AgCl+Ca\left(NO_3\right)_2\)
\(\rightarrow n_{AgBr}=x\left(mol\right);n_{AgCl}=2y\left(mol\right)\)
\(\rightarrow m_{kt}=2m=188x+287y\)
\(\rightarrow188x+287y=2\left(103x+111y\right)\)
\(\rightarrow65y=18x\rightarrow x=\frac{65y}{18}\)
\(\rightarrow\%m_{NaBr}=\frac{\frac{103.65y}{18}}{\frac{103.65y}{18+111y}}=77\%\)
\(\rightarrow\%m_{CaCl2}=100\%-77\%=23\%\)