a) 4Na + O2 --to--> 2Na2O
\(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)
Theo PTHH: \(n_{Na_2O}=0,05\left(mol\right)\)
=> \(m_{Na_2O}=0,05.62=3,1\left(g\right)\)
b)
PTHH: Na2O + H2O --> 2NaOH
0,05-------------->0,1
=> mNaOH = 0,1.40 = 4 (g)
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