Câu 1:
\(n_{Na_2O}=\dfrac{6,2}{62}=0,1mol\)
Na2O+H2O\(\rightarrow\)2NaOH
\(n_{NaOH}=2n_{Na_2O}=0,2mol\)
\(C\%_{NaOH}=\dfrac{0,2.40.100}{100+6,2}\approx7,53\%\)
Câu 2:
\(n_{Na}=\dfrac{6,2}{23}mol\)
2Na+2H2O\(\rightarrow\)2NaOH+H2
Số mol NaOH=số mol Na=\(\dfrac{6,2}{23}mol\)
Số mol H2=0,5 số mol Na=\(\dfrac{3,1}{23}mol\)
mdd=6,2+100-\(\dfrac{3,1}{23}.2\approx105,93g\)
\(C\%_{NaOH}=\dfrac{\dfrac{6,2}{23}.40.100}{105,93}\approx10,2\%\)