Đáp án:
\(\begin{array}{l} C\%_{NaOH}\text{(dư)}=2,01\%\\ C\%_{Na_2SO_4}=3,57\%\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l} PTHH:\\ 2Na+2H_2O\to 2NaOH+H_2↑\ (1)\\ 2NaOH+CuSO_4\to Na_2SO_4+Cu(OH)_2↓\ (2)\\ n_{Na}=\dfrac{9,2}{23}=0,4\ mol.\\ Theo\ pt\ (1):\ n_{NaOH}=n_{Na}=0,4\ mol.\\ n_{CuSO_4}=\dfrac{400\times 4\%}{160}=0,1\ mol.\\ \text{Lập tỉ lệ:}\ \dfrac{0,4}{2}>\dfrac{0,1}{1}\\ ⇒NaOH\ \text{dư.}\\ Theo\ pt\ (1):\ n_{H_2}=\dfrac{1}{2}n_{Na}=0,2\ mol.\\ Theo\ pt\ (2):\ n_{NaOH}\text{(pư)}=2n_{CuSO_4}=0,2\ mol.\\ ⇒n_{NaOH}\text{(dư)}=0,4-(0,1\times 2)=0,2\ mol.\\ Theo\ pt:\ n_{Cu(OH)_2}=n_{Na_2SO_4}=n_{CuSO_4}=0,1\ mol.\\ ⇒m_{\text{dd spư}}=0,2\times 40+400-0,1\times 98-0,2\times 2=397,8\ g.\\ ⇒C\%_{NaOH}\text{(dư)}=\dfrac{0,2\times 40}{397,8}\times 100\%=2,01\%\\ C\%_{Na_2SO_4}=\dfrac{0,1\times 142}{397,8}\times 100\%=3,57\%\end{array}\)
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