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a, \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PT: \(2Na+2H_2O\rightarrow2NaOH+H_2\)
Theo PT: \(n_{Na}=2n_{H_2}=0,4\left(mol\right)\Rightarrow m_{Na}=0,4.23=9,2\left(g\right)\)
b, \(n_{NaOH}=2n_{H_2}=0,4\left(mol\right)\)
\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,4}{2}=0,2\left(M\right)\)
\(n_{Fe}=\dfrac{11,2}{56}=0,2mol\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
0,2 0,2 0,2 0,2
a)\(V_{H_2}=0,2\cdot22,4=4,48l\)
b)\(C_{M_{H_2SO_4}}=\dfrac{0,2}{0,5}=0,4M\)
c)\(C_{M_{FeSO_4}}=\dfrac{0,2}{0,5}=0,4M\)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
pthh : \(Fe+H_2SO_4->FeSO_4+H_2\)
0,2 0,2
=> \(V_{H_2}=0,2.22,4=4,48\left(L\right)\)
\(m_{H_2SO_4}=\dfrac{0,5}{22,4}.98\approx2,188\left(g\right)\)
=> mdd=11,2+2,188=13,388(g)
C%=\(\dfrac{2,188}{13,388}.100\%=16,3\%\)
\(a,n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\)
PTHH: Na2O + H2O ---> 2NaOH
0,2----------------->0,4
=> mNaOH = 0,4.40 = 16 (g)
b) mdd = 12,4 + 27,6 = 40 (g)
=> \(C\%_{NaOH}=\dfrac{16}{40}.100\%=40\%\)
\(n_{Na}=\dfrac{4.6}{23}=0.2\left(mol\right)\)
\(4Na+O_2\underrightarrow{^{^{t^0}}}2Na_2O\)
\(0.2...................0.1\)
\(Na_2O+H_2O\rightarrow2NaOH\)
\(0.1.......................0.2\)
Quỳ tím hóa xanh => Vì dung dịch NaOH có tính bazo
\(m_{dd_A}=\dfrac{m_{NaOH}}{8\%}=\dfrac{0.2\cdot40}{8\%}=100\left(g\right)\)
\(n_{Al_2O_3}=\dfrac{5.1}{102}=0.05\left(mol\right)\)
\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(0.05...........0.15...............0.05\)
\(C_{M_{H_2SO_4}}=\dfrac{0.15}{0.2}=0.75\left(M\right)\)
\(C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0.05}{0.2}=0.25\left(M\right)\)
2Na + 2H2O \(\rightarrow\) 2NaOH + H2 (1)
K2O + H2O \(\rightarrow\) 2KOH (2)
Có : mdd B = mhh A + mH2O - mH2 = 19,85 + 180,4 - mH2 = 200
\(\Rightarrow\) mH2 = 0,25(g)
\(\Rightarrow\) nH2 = 0,25/2 = 0,125(mol)
Theo PT(1) \(\Rightarrow\)nNa = nNaOH = 2.nH2 = 2. 0,125 = 0,25(mol)
\(\Rightarrow\left\{{}\begin{matrix}m_{Na}=0,25.23=5,75\left(g\right)\\m_{NaOH}=0,25.40=10\left(g\right)\end{matrix}\right.\)
\(\Rightarrow\) mK2O = 19,85 - 5,75= 14,1(g)
\(\Rightarrow\) nK2O = 14,1/94 = 0,15(mol)
Theo PT(2) \(\Rightarrow\) nKOH = 2 . nK2O = 2. 0,15 = 0,3(mol)
\(\Rightarrow\) mKOH = 0,3 . 56 = 16,8(g)
* C%KOH / ddB = 16,8/200 . 100% = 8,4%
C%NaOH / dd B = 10/200 . 100% = 5%
* m(KOH+ NaOH) = 16 ,8 + 10 =\ 26,8(g)
\(\Rightarrow\)mH2O / dd B = 200 - 26,8 = 173,2 (g)
\(\Rightarrow\) VH2O / dd B = m : D = 173,2 : 1 = 173,2 (ml) =0,1732(l)
mà Vdd B = VH2O / ddB
=> Vdd B =\ 0,1732(l)
Do đó :
CM của NaOH / dd B = 0,25/0,1732=1,44(M)
CM của KOH / dd B = 0,3/0,1732 = 1,73 (M)
$n_{Na_2O} = \dfrac{12,4}{62} = 0,2(mol)$
$n_{Na} = \dfrac{4,6}{23} = 0,2(mol)$
$n_{NaOH} = 1.V = V(mol)$
$Na_2O + H_2O \to 2NaOH$
$2Na + 2H_2O \to 2NaOH + H_2$
$n_{NaOH\ tạo thành} = 2n_{Na_2O} + n_{Na} = 0,6(mol)$
$\Rightarrow 0,6 + V = V.5$
$\Rightarrow V = 0,15(lít)$