Đáp án:
\(b,\ C\%_{H_2SO_4}=4,083\%\\ c,\ C\%_{Na_2SO_4}=3,55\%\)
Giải thích các bước giải:
\(a,\ PTHH:2NaOH+H_2SO_4\to Na_2SO_4+2H_2O\\ b,\ m_{NaOH}=200\times 5\%=10\ g.\\ ⇒n_{NaOH}=\dfrac{10}{40}=0,25\ mol.\\ Theo\ pt:\ n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,125\ mol.\\ ⇒C\%_{H_2SO_4}=\dfrac{0,125\times 98}{300}\times 100\%=4,083\%\\ c,\ m_{\text{dd spư}}=m_{\text{dd NaOH}}+m_{\text{dd H$_2$SO$_4$}}=200+300=500\ g.\\ Theo\ pt:\ n_{Na_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,125\ mol.\\ ⇒C\%_{Na_2SO_4}=\dfrac{0,125\times 142}{500}\times 100\%=3,55\%\)
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