PTHH: \(2NaCl+2H_2O\underrightarrow{đp}2NaOH+Cl_2+H_2\)
Đổi \(22,4m^3=22400\left(l\right)\)
Ta có: \(V_{Cl_2\left(líthuyết\right)}=\dfrac{22400}{80\%}=28000\left(l\right)\) \(\Rightarrow n_{Cl_2}=\dfrac{28000}{22,4}=1250\left(mol\right)\)
\(\Rightarrow n_{NaCl}=n_{NaOH}=2500\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=2500\cdot58,5=146250\left(g\right)=146,25\left(kg\right)\\m_{ddNaOH}=\dfrac{2500\cdot40}{40\%}=250000\left(g\right)=250\left(kg\right)\end{matrix}\right.\)
Đúng 2
Bình luận (0)
