Câu 29:
a, Ta có: \(n_{NO}=0,02\left(mol\right)\)
Theo ĐLBT e, có: \(3n_{Fe}=3n_{NO}\Rightarrow n_{Fe}=0,02\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,02.56}{1,52}.100\%\approx73,68\%\\\%m_{MgO}\approx26,32\%\end{matrix}\right.\)
b, Ta có: \(n_{MgO}=\dfrac{1,52-0,02.56}{40}=0,01\left(mol\right)\)
BTNT Fe: \(n_{Fe\left(NO_3\right)_3}=n_{Fe}=0,02\left(mol\right)\)
BTNT Mg: \(n_{Mg\left(NO_3\right)_2}=n_{MgO}=0,01\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C_{M_{Fe\left(NO_3\right)_3}}=\dfrac{0,02}{0,2}=0,1\left(M\right)\\C_{M_{Mg\left(NO_3\right)_2}}=\dfrac{0,01}{0,2}=0,05\left(M\right)\end{matrix}\right.\)
BTNT N, có: \(n_{HNO_3\left(pư\right)}=3n_{Fe\left(NO_3\right)_3}+2n_{Mg\left(NO_3\right)_2}+n_{NO}=0,1\left(mol\right)\)
Mà: \(n_{HNO_3\left(banđau\right)}=0,2.1=0,2\left(mol\right)\)
\(\Rightarrow n_{HNO_3\left(dư\right)}=0,1\left(mol\right)\Rightarrow\left[H^+\right]=\dfrac{0,1}{0,2}=0,5\left(M\right)\)
\(\Rightarrow pH=-\log\left(0,5\right)\approx0,3\)
c, \(H_{\left(dư\right)}^++OH^-\rightarrow H_2O\)
_0,1______0,1 (mol)
\(Fe^{3+}+3OH^-\rightarrow Fe\left(OH\right)_3\)
0,02_____0,06 (mol)
\(Mg^{2+}+2OH^-\rightarrow Mg\left(OH\right)_2\)
0,01_____0,02 (mol)
\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,1+0,06+0,02}{0,2}=0,9\left(M\right)\)