PTHH: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
Ta có: \(n_{NaOH}=\dfrac{0,8}{40}=0,02\left(mol\right)\\ =>n_{Na_2SO_4}=\dfrac{0,02}{2}=0,01\left(mol\right)\)
=>\(m_{muối}=142.0,01=1,42\left(g\right)\)
\(\left\{{}\begin{matrix}m_{NaOH}=0,8gam;M_{NaOH}=23+16+1=40\\SómolNaOH.n_{NaOH}=\dfrac{m}{M}=\dfrac{0,8}{40}=0,02mol\end{matrix}\right.\)
PT: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
\(0,02mol\rightarrow\dfrac{0,02}{2}=0,01mol\)
\(\left\{{}\begin{matrix}n_{Na_2SO_4}=0,01mol;M_{Na_2SO_4}=23.2+32+16.4=142\\\Rightarrow khốilượngNa_2SO_4.m_{Na_2SO_4}=n.M=0,01.142=1,42g\end{matrix}\right.\)
Vậy khối lượng muối Na2SO4 khan là 1,42gam
