2Na + 2H2O → 2NaOH + H2
\(m_{Na}=2\times23=46\left(g\right)\)
\(m_{H_2O}=200\times1=200\left(g\right)\)
Theo pT: \(n_{H_2}=\frac{1}{2}n_{Na}=\frac{1}{2}\times2=1\left(mol\right)\)
\(\Rightarrow m_{H_2}=1\times2=2\left(g\right)\)
Ta có: \(m_{ddA}=46+200-2=244\left(g\right)\)
Theo pT: \(n_{NaOH}=n_{Na}=2\left(mol\right)\)
\(\Rightarrow m_{NaOH}=2\times40=80\left(g\right)\)
\(\Rightarrow C\%_{NaOH}=\frac{80}{244}\times100\%=32,79\%\)
PTHH : \(Na+H_2O\rightarrow NaOH+\frac{1}{2}H_2\uparrow\) (1)
Theo PTHH (1) : \(n_{Na}=2mol;n_{NaOH}=2mol;n_{H_2}=1mol\)
Có : \(m_{H_2O}=V_{H_2O}.D_{H_2O}=200.1=200g\)
Vì \(n_{Na}=2< n_{H_2O}=\frac{200}{18}\) : Sau p/ứng (1) thì H2O dư
Khối lượng dung dịch thu được :
\(m_{dd}=m_{Na}+m_{H_2O}-m_{H_2}=2.23+200-1.2=244g\)
Từ (1) => \(n_{NaOH}=2mol\Rightarrow m_{NaOH}=2.40=80g\)
Vậy \(C\%_{NaOH}=\frac{80}{244}.100=32,79\%\)
mH2O = 200 g
nH2O = 11.11 mol
2Na + 2H2O --> 2NaOH + H2
2______2________2______1
mdd sau phản ứng = 46 + 200 - 2 = 244 g
mNaOH = 2*40 = 80g
C%NaOH = 80/244*100% = 32.78%
