\(n_{Na_2O}=\dfrac{31}{62}=0.5\left(mol\right)\)
\(Na_2O+H_2O\rightarrow2NaOH\)
\(0.5........................1\)
\(m_{NaOH}=1\cdot40+500\cdot10\%=90\left(g\right)\)
\(m_{dd_{NaOH\left(sau\right)}}=31+500=531\left(g\right)\)
\(C\%_{NaOH}=\dfrac{90}{531}\cdot100\%=16.9\%\)
\(n_{Na_2O}=\dfrac{31}{62}=0,5\left(mol\right)\)
\(n_{NaOH\left(bd\right)}=\dfrac{500.10}{100.40}=1,25\left(mol\right)\)
PTHH: Na2O + H2O --> 2NaOH
_______0,5----------------->1__________(mol)
nNaOH(sau pư) = 1,25 + 1 = 2,25 (mol)
\(C\%_{NaOH}=\dfrac{2,25.40}{31+500}.100\%=16,95\%\)
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https://khoahoc.vietjack.com/question/474748/cho-31-gam-na2o-vao-500-gam-dung-dich-naoh-10
