\(n_{NaOH}=\frac{m}{M}=\frac{20}{40}=0,5mol\)
PTHH:
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
0,5 0,25 0,25 0,5 (mol)
b)\(m_{H_2SO_4}=n.M=0,25.98=24,5g\)
\(m_{ddH_2SO_4}=\frac{24,5.100}{20}=122,5g\)
c)\(m_{Na_2SO_4}=n.M=0,25.142=35,5g\)\(m_{ddNa_2SO_4}=m_{NaOH}+m_{ddH_2SO_4}=20+122,5=142,5g\)\(C\%_{Na_2SO_4}=\frac{35,5.100}{142,5}=24,91\%\)
PTHH: 2NaOH+H2SO4--->Na2SO4+2H2O
nNaOH = 20/40 = 0,5 (mol)
=>mH2SO4 = 0,25.98 = 24, 5 g
=> mddH2SO4 = 122,5g
=>mNa2SO4 = 0,25. 142 = 35,5g
=> mdd Na2SO4 = 142,5g
=>C%Na2SO4 = 24,91%
2NaOH + H2SO4---->Na2SO4+ 2H2O
a) Ta có
n\(_{NaOH}=\frac{20}{40}=0,5\left(mol\right)\)
Theo pthh
n\(_{H2SO4}=\frac{1}{2}n_{NaOH}=0,25\left(mol\right)\)
m\(_{H2SO4}=\frac{0,25.98.100}{20}=122,5\left(g\right)\)
b)Theo pthh
n\(_{Na2SO4}=\frac{1}{2}n_{NaOH}=0,25\left(mol\right)\)
m\(_{Na2SO4}=0,25.142=35,5\left(g\right)\)
C%=\(\frac{35,5}{122,5+40}.100\%=21,85\%\)
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