n\(_{Na}\)= \(\dfrac{6,9}{23}\)= 0,3 (mol)
m\(_{NaOH}\) = 150 . \(\dfrac{20}{100}\) = 30 (g)
PTHH: 2Na + 2H2O ----> 2NaOH + H2
mol: ....0,3------------------>0,3
\(\sum\)m\(_{NaOH}\) = 30 + 0,3 . 40 = 42 (g)
mdd\(_{NaOH}\) = 150 + 6,9 = 156,9 (g)
C% NaOH = \(\dfrac{42}{156,9}\).100% = 26,77%
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