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\(C\%_{NaOH}=\dfrac{a}{a+40}=20\%\\ \Leftrightarrow a=10\left(g\right)\)
\(m_{NaCl\left(10\%\right)}=10\%.b=0,1b\left(g\right)\\ m_{NaCl\left(20\%\right)}=0,1b+10\left(g\right)\\ \rightarrow C\%_{NaCl\left(20\%\right)}=\dfrac{0,1b+10}{b}=20\%\\ \Leftrightarrow b=100\left(g\right)\)
Bài 1:
\(a.n_{NaOH\left(tổng\right)}=0,05.1+0,2.0,2=0,09\left(mol\right)\\ V_{ddNaOH\left(tổng\right)}=50+200=250\left(ml\right)=0,25\left(l\right)\\ C_{MddNaOH\left(cuối\right)}=\dfrac{0,09}{0,25}=0,36\left(M\right)\\ b.n_{HCl}=0,5.0,02=0,01\left(mol\right)\\ n_{H_2SO_4}=0,08.0,2=0,016\left(mol\right)\\ V_{ddsau}=20+80=100\left(ml\right)=0,1\left(l\right)\\ C_{MddH_2SO_4}=\dfrac{0,016}{0,1}=0,16\left(M\right)\\ C_{MddHCl}=\dfrac{0,01}{0,1}=0,1\left(M\right)\)
Bài 2:
\(a.m_{H_2SO_4}=29,4.10\%=2,94\left(g\right)\\ b.n_{H_2SO_4}=\dfrac{2,94}{98}=0,03\left(mol\right)\\ n_{Fe}=\dfrac{0,56}{56}=0,01\left(mol\right)\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ Vì:\dfrac{0,01}{1}< \dfrac{0,03}{1}\Rightarrow H_2SO_4dư\\ n_{H_2SO_4\left(dư\right)}=0,03-0,01=0,02\left(mol\right)\\ m_{H_2SO_4\left(dư\right)}=0,02.98=1,96\left(g\right)\\ n_{H_2}=n_{Fe}=0,01\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,01.22,4=0,224\left(l\right)\)
\(SO_3+H_2O\rightarrow H_2SO_4\\ m_{H_2SO_4\left(tăng\right)}=\dfrac{98}{80}m=\dfrac{49}{40}m=1,225m\left(g\right)\\ m_{H_2SO_4\left(dd.14,7\%\right)}=14,7.200=29,4\left(g\right)\\ Ta.có:C\%_{ddH_2SO_4\left(cuối\right)}=20\%\\ \Leftrightarrow\dfrac{29,4+1,225m}{m+200}.100\%=20\%\\ m\approx10,3415\left(g\right)\)
\(m_{H_2SO_4}=\dfrac{20.a}{100}=0,2a\left(g\right)\)
mdd sau khi thêm = 40 + 20 = 60 (g)
\(C\%_{dd.sau.khi.thêm}=\dfrac{0,2a}{60}.100\%=10\%\)
=> a = 30