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PTHH: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
Ta có: \(n_{NaOH}=\dfrac{0,8}{40}=0,02\left(mol\right)\\ =>n_{Na_2SO_4}=\dfrac{0,02}{2}=0,01\left(mol\right)\)
=>\(m_{muối}=142.0,01=1,42\left(g\right)\)
\(\left\{{}\begin{matrix}m_{NaOH}=0,8gam;M_{NaOH}=23+16+1=40\\SómolNaOH.n_{NaOH}=\dfrac{m}{M}=\dfrac{0,8}{40}=0,02mol\end{matrix}\right.\)
PT: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
\(0,02mol\rightarrow\dfrac{0,02}{2}=0,01mol\)
\(\left\{{}\begin{matrix}n_{Na_2SO_4}=0,01mol;M_{Na_2SO_4}=23.2+32+16.4=142\\\Rightarrow khốilượngNa_2SO_4.m_{Na_2SO_4}=n.M=0,01.142=1,42g\end{matrix}\right.\)
Vậy khối lượng muối Na2SO4 khan là 1,42gam
PTHH: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
Ta có: \(n_{Fe_2O_3}=\dfrac{4,8}{160}=0,03\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{H_2SO_4}=0,09\left(mol\right)\\n_{Fe_2\left(SO_4\right)_3}=0,03\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddH_2SO_4}=\dfrac{0,09\cdot98}{9,8\%}=90\left(g\right)\\m_{Fe_2\left(SO_4\right)_3}=0,03\cdot400=12\left(g\right)\end{matrix}\right.\)
\(n_{FeCl_3}=\dfrac{48,75}{162,5}=0,3(mol)\\ 3NaOH+FeCl_3\to Fe(OH)_3\downarrow+3NaCl\\ \Rightarrow n_{Fe(OH)_3}=0,3(mol);n_{NaOH}=n_{NaCl}=0,9(mol)\\ a,m_{Fe(OH)_3}=0,3.107=32,1(g)\\ b,m_{dd_{NaOH}}=\dfrac{0,9.40}{10\%}=360(g)\\ c,C\%_{NaCl}=\dfrac{0,9.58,5}{360+48,75-32,1}.100\%=13,98\%\\ \)
\(d,2Fe(OH)_3+3H_2SO_4\to Fe_2(SO_4)_3+6H_2O\\ \Rightarrow n_{H_2SO_4}=0,45(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,45.98}{20\%}=220,5(g)\\ \Rightarrow V_{dd_{H_2SO_4}}=\dfrac{220,5}{1,14}=193,42(ml)\)
\(n_{NaOH}=\dfrac{0,8}{40}=0,02mol\\ 2NaOH+H_2SO_4->Na_2SO_4+2H_2O\\ m_{Na_2SO_4}=142\cdot0,01=1,42g\\ n_{H_2SO_4pư}=0,01mol\\ m_{H_2SO_4}=98\cdot1,15\cdot0,01=1,127g\)
\(a.n_{NaOH}=\dfrac{0,8}{40}=0,02\left(mol\right)\\2 NaOH+H_2SO_4\xrightarrow[]{}Na_2SO_4+2H_2O\\ \Rightarrow n_{Na_2SO_4}=\dfrac{1}{2}0,02=0,01\left(mol\right)\\ m_{Na_2SO_4}=0,01.142=1,42\left(g\right)\\ b.n_{H_2SO_4\left(pư\right)}=\dfrac{1}{2}0,02=0,01\left(mol\right)\\ n_{H_2SO_4\left(dư\right)}=0,01.15\%=0,0015\left(mol\right)\\ m_{H_2SO_4\left(dùng\right)}=\left(0,01+0,0015\right).98=1,127\left(g\right)\)