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\(n_{Zn\left(OH\right)_2\left(mp\right)}=\dfrac{19.8}{99\cdot2}=0.1\left(mol\right)\)
\(a.\)
\(n_{H_2SO_4}=0.15\left(mol\right)\)
\(Zn\left(OH\right)_2+H_2SO_4\rightarrow ZnSO_4+2H_2O\)
Lập tỉ lệ :
\(\dfrac{0.1}{1}< \dfrac{0.15}{1}\) \(\Rightarrow H_2SO_4dư\)
\(m_{ZnSO_4}=0.1\cdot161=16.1\left(g\right)\)
\(b.\)
\(n_{NaOH}=0.15\left(mol\right)\)
\(2NaOH+Zn\left(OH\right)_2\rightarrow Na_2ZnO_2+2H_2O\)
Lập tỉ lệ :
\(\dfrac{0.15}{2}< \dfrac{0.1}{1}\Rightarrow Zn\left(OH\right)_2dư\)
\(n_{Na_2ZnO_2}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}\cdot0.15=0.075\left(mol\right)\)
\(m=0.075\cdot143=10.725\left(g\right)\)
a, (1) 4P+5.O2->2.P2O5
(2) P2O5+4.NaOH->2.Na2HPO4+H2O
b, photpho có n=6,2:31=0,2 mol.dựa theo pt (1) thấy nP2O5=0,1mol.theo pt (2) thấy nNaOH=0,4mol vậy mNaOH=0,4.40=16 g vậy m(dd NaOH)=16:32%=50 g
c, theo pt (2) nNa2HPO4 =0,2 mol vậy mNa2HPO4=0,2.142=28,4 g
m(dd sau pư)=mP+m(dd NaOH)=6,2+50=56,2 g
=> C%(dd Na2HPO4)=28,4:56,2=50,53%
a/
\(n_{KOH}=0,1.0,2=0,02\left(mol\right)\)
\(KOH\rightarrow K^++OH^-\)
0,02 0,02 0,02 (mol)
vì dung dịch trung hòa nên:
\(n_{OH^-}=n_{H^+}=0,02\left(Mol\right)\)
\(H_2SO_4\rightarrow2H^++SO_4^{2-}\)
0,01 0,02 0,01 (mol)
\(C_{M_{H_2SO_4}}=\dfrac{0,01}{0,1}=0,1\left(M\right)\)
\(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
0,2 0,1 0,1 (mol)
\(m_{K_2SO_4}=0,1.174=17,4\left(g\right)\)
b/
vì dung dịch trung hòa nên:
\(n_{H^+}=n_{OH^-}=0,2\left(mol\right)\)
\(HCl\rightarrow H^++Cl^-\)
0,2 0,2 0,2 (mol)
\(KOH+HCl\rightarrow KCl+H_2O\)
0,2 0,2 0,2 (mol)
\(m_{KCl}=0,2.74,2=14,9\left(g\right)\)
a) PI : C3H6 +Br2 --> C3H6Br2 (1)
C3H4 +2Br2 --> C3H4Br4 (2)
PII : C3H4 + AgNO3 +NH3 --> C3H3Ag\(\downarrow\) + NH4NO3
nC3H3Ag=0,04(mol)
=> nC3H4=0,04(mol)=> nC3H4( hh đầu)=0,08(mol)
Theo (1,2) : nB2=2nC3H4+nC3H6=0,11
=> nC3H6=0,03(mol) => nC3H6(hh đầu)=0,06(mol)
=>nC2H6(bđ)=0,02(mol)
=> %mC2H6=9,49(%)
%mC3H4=50,63(%)
%mC3H6=39,87(%)
b) C2H6-->2CO2
0,02 --> 0,04
C3H4 --> 3CO2
0,08 --> 0,24
C3H6 --> 3CO2
0,06 --> 0,18
=> \(\Sigma\)nCO2(tạo ra)=0,46(mol)
CO2 + Ba(OH)2 --> BaCO3 +H2O (3)
CO2 + BaCO3 +H2O --> Ba(HCO3)2 (4)
Tho (3,4) : nBaCO3(tạo ra)=0,18(mol)
=> mBaCO3=35,46(g)
Muối thu đk:: Na+ :0,3mol , K+ :0,1mol , SO42- :0,2 mol → m=30g
nAl(OH)3= 15.6/78=0.2 mol
PI : nH2SO4= 0.2 mol
2Al(OH)3 + 3H2SO4 --> Al2(SO4)3 + 6H2O
Bđ: 0.2_________0.2
Pư : 1/15________0.2_________1/15
Kt: 2/15_________0___________1/15
mAl2(SO4)3= 1/15*342= 22.8g
PII:
nNaOH = 0.05 mol
NaOH + Al(OH)3 --> NaAlO2 + 2H2O
Bđ: 0.05_____0.2
Pư: 0.05_____0.05______0.05
Kt: 0________0.15______0.05
mNaAlO2= 0.05*82=4.1g
a) \(n_{H_2SO_4}=0,2\times1=0,2\left(mol\right)\)
\(m_{Al\left(OH\right)_3}=\frac{1}{2}\times15,6=7,8\left(g\right)\)
\(\Rightarrow n_{Al\left(OH\right)_3}=\frac{7,8}{78}=0,1\left(mol\right)\)
PTHH: 2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O
Theo PT: \(n_{Al\left(OH\right)_3}=\frac{2}{3}n_{H_2SO_4}\)
Theo bài: \(n_{Al\left(OH\right)_3}=\frac{1}{2}n_{H_2SO_4}\)
Vì \(\frac{1}{2}< \frac{2}{3}\) ⇒ H2SO4 dư
Theo PT: \(n_{Al_2\left(SO_4\right)_3}=\frac{1}{2}n_{Al}=\frac{1}{2}\times0,1=0,05\left(mol\right)\)
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,05\times342=17,1\left(g\right)\)
b) \(n_{NaOH}=0,05\times1=0,05\left(mol\right)\)
\(m_{Al\left(OH\right)_3}=\frac{1}{2}\times15,6=7,8\left(g\right)\)
\(\Rightarrow n_{Al\left(OH\right)_3}=\frac{7,8}{78}=0,1\left(mol\right)\)
PTHH: Al(OH)3 + NaOH → NaAlO2 + 2H2O
Theo PT: \(n_{Al\left(OH\right)_3}=n_{NaOH}\)
Theo bài : \(n_{Al\left(OH\right)_3}=2n_{NaOH}\)
Vì \(2>1\) → Al(OH)3 dư
Theo Pt: \(n_{NaAlO_2}=n_{NaOH}=0,05\left(mol\right)\)
\(\Rightarrow m_{NaAlO_2}=0,05\times82=4,1\left(g\right)\)