Bài 4:
a) Coi nước là dd NaCl có C% = 0
Áp dụng sơ đồ đường chéo:
\(\dfrac{m_{ddNaCl\left(10\%\right)}}{m_{H_2O}}=\dfrac{10-5}{5-0}=\dfrac{1}{1}\)
=> \(n_{H_2O}=n_{NaCl}=200\left(g\right)\)
b) Coi nước là dd H2SO4 có C% = 0
Áp dụng sơ đồ đường chéo, ta có:
\(\dfrac{m_{H_2O}}{m_{ddH_2SO_4\left(12\%\right)}}=\dfrac{12-8}{8-0}=\dfrac{1}{2}\)
Mà \(m_{H_2O}+m_{ddH_2SO_4\left(12\%\right)}=m_{ddH_2SO_4\left(8\%\right)}=400\left(g\right)\)
=> \(m_{H_2O}=400.\dfrac{1}{1+2}\approx133,33\left(g\right)\)
Bài 5:
Gọi \(\left\{{}\begin{matrix}n_{H_2SO_4}=a\left(mol\right)\\n_{H_2O}=b\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\sum n_O=4n_{H_2SO_4}+n_{H_2O}=4a+b\left(mol\right)\\\sum n_O=2n_{H_2SO_4}+2n_{H_2O}=2a+2b\left(mol\right)\end{matrix}\right.\)
Mà: \(\dfrac{n_O}{n_H}=1,25\)
`=>` \(\dfrac{4a+b}{2a+2b}=1,25\)
`<=>` \(4a+b=2,5a+2,5b\)
`<=>` \(1,5a=1,5b\)
`<=> a = b`
`=>` \(n_{H_2SO_4}=n_{H_2O}\)
`=>` \(C\%_{H_2SO_4}=\dfrac{98a}{98a+18b}.100\%=\dfrac{98a}{98a+18b}.100\%=84,48\%\)
