Ta co pthh
NaOH + HCl \(\rightarrow\) NaCl + H2O
Theo de bai ta co
Khoi luong cua NaOH trong 400g dd NaOH 30% la
mNaOH =\(\dfrac{mdd.C\%}{100\%}=\dfrac{400.30\%}{100\%}=120g\)
\(\Rightarrow\) nNaOH=\(\dfrac{120}{40}=3mol\)
a, Theo pthh
nNaCl=nNaOH=3mol
\(\Rightarrow\) mNaCl=3.58,5=175,5 g
mdd NaCl=mNaOH + mddHCl-mH2O=400+100-(3.18)=446g
\(\Rightarrow\) C%=\(\dfrac{mct}{mdd}.100\%=\dfrac{175,5}{446}.100\%\approx39,35\%\)
\(NaOH(3)+HCl(3)--->NaCl(3)+H_2O\)
\(m_{NaOH}=120\left(g\right)\)
\(\Rightarrow n_{NaOH}=3\left(mol\right)\)
\(a)\)
Muối thu được sau phản ứng là NaCl
Theo PTHH: \(n_{NaCl}=3\left(mol\right)\)
\(\Rightarrow m_{NaCl}=175,5\left(g\right)\)
\(m dd sau = 400+100=500(g)\)
\(C\%_{NaCl}=\dfrac{175,5}{500}.100\%=35,1\%\)
\(b)\)
Đề thiếu. Sửa đề: Tính nồng độ phần trăm của dung dich axit HCl đã dùng biết pứ xảy ra hoàn toàn.
Theo PTHH: \(n_{HCl}=3\left(mol\right)\)
\(\Rightarrow C\%_{HCl}=\dfrac{3.36,5}{100}.100\%=109,5\%\)
ta có pthh:NaOH+HCL\(\rightarrow\)H\(_2\)O+NaCL
theo gt ta có: m\(_{NaCL}\)=(C\(\)%\(\times\)m\(_{d_{ }d\dfrac{ }{ }}\))\(\div\)100
=(30\(\times\)400)\(\div\)100
=120(g)
a) ta có:
nNaOH=\(\)3(mol)
theo pthh ta có n\(_{Nacl=}\)n\(_{NaOH}\)
=3mol
\(\Rightarrow\) m\(_{NaCL}\)=3\(\times\)58.5
=175.5(g)
\(\Rightarrow\)C%= \(\dfrac{175.5}{446}\) \(\times\)100
=39.35%
b) ta có:(3\(\times\)36.5)\(\div\)100\(\times\)100
=109.5
5
\(\Rightarrow\)C%=\(\dfrac{m_{ct}}{m_d_d}\)
=
GIÚP VỚI Ạ MAI EM THI HK 2 RÙI