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\(PTHH:2Na+2H_2O\rightarrow2NaOH+H_2\)
\(n_{Na}=\dfrac{46}{23}=2\left(mol\right)\)
Theo pthh:
\(\left\{{}\begin{matrix}n_{H_2}=\dfrac{1}{2}n_{Na}=1\left(mol\right)\\n_{NaOH}=n_{Na}=2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{H_2}=1.2=2\left(g\right)\\m_{NaOH}=2.40=80\left(g\right)\end{matrix}\right.\)
Theo đlbtkl, ta có:
\(m_{ddNaOH}=m_{Na}+m_{H_2O}-m_{H_2}\)
\(=46+180-2=224\left(g\right)\)
\(C\%ddNaOH=\dfrac{80}{224}.100\approx35,7\%\)
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: \(n_{NaOH}=n_{Na}=2\left(mol\right)\)
\(\Rightarrow m_{NaOH}=2\times40=80\left(g\right)\)
\(\Rightarrow C\%_{NaOH}=\frac{80}{244}\times100\%=32,79\%\)
nNa=46/23=2 mol
2Na +1/2 O2 =>Na2O
2 mol =>0,5 mol=>1 mol
Na2O + H2O =>2NaOH
1 mol =>2 mol
mdd A=mNa2O+mH2O=1.62+400=462g
nNaOH=2 mol
=>mNaOH=2.40=80g
=>C% dd NaOH=80/462.100%=17,32%
CM dd NaOH=CM dd A=2/0,4=5M
\(n_{Na}=\frac{46}{23}=2\left(mol\right)\)
\(m_{H_2O}=1.400=400\left(g\right)\)
\(4Na_{ }+O_{2_{ }}\rightarrow2Na_2O\)
2mol 1mol
\(Na_2O+H_2O\rightarrow2NaOH\)
1mol 2mol
\(m_{NaOH}=2.40=80\left(g\right)\)
Khối lượng dung dịch sau phản ứng là:
\(m_{d_2}=46+400=446\left(g\right)\)
\(C\%=\frac{80}{446}.100\%=17,94\%\)
Đổi 400ml = 0,4l
\(C_M=\frac{2}{0,4}=5\left(M\right)\)
\(n_{K_2O}=\dfrac{21,15}{94}=0,225\left(mol\right)\\
pthh:K_2O+H_2O\rightarrow KOH+\dfrac{1}{2}H_2\)
0,225 0,225 0,1125
\(m_{\text{dd}}=21,15+178,85-\left(0,1125.2\right)=199,775g\\
C\%=\dfrac{0,225.56}{199,775}.100\%=6,3\%\)
c)
\(m_{\text{dd}}=21,15+\left(50+178,85\right)-\left(0,1125.2\right)=249,775g\\
C\%=\dfrac{0,225.56}{249,775}.100\%=5\%\)
\(n_{Na}=\dfrac{46}{23}=2\left(mol\right)\)
\(2Na+2H_2O\rightarrow2NaOH+H_2\)
\(2..........................2................1\)
\(m_A=m_{Na}+m_{H_2O}-m_{H_2}=46+200-2=244\left(g\right)\)
\(m_{NaOH}=2\cdot40=80\left(g\right)\)
\(C\%NaOH=\dfrac{80}{244}\cdot100\%=32.78\%\)
Đề cho thế này sao tính đc C% nhỉ