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Gọi nNa2CO3 = x (mol)
Na2CO3 + 2HCl \(\rightarrow\) 2NaCl + H2O + CO2
x \(\rightarrow\) 2x \(\rightarrow\) 2 x (mol)
C%(NaCl) = \(\frac{2.58,5x}{200+120}\) . 100% = 20%
=> x =0,547 (mol)
mNa2CO3 = 0,547 . 106 = 57,982 (g)
mHCl = 2 . 0,547 . 36,5 =39,931 (g)
C%(Na2CO3) =\(\frac{57,892}{200}\) . 100% = 28,946%
C%(HCl) = \(\frac{39,931}{120}\) . 100% = 33,28%
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)
nH2=4,48/22,4=0,2(mol)
=>nFe=0,2(mol)=>mFe=0,2.56=11,2(g)
=>mFeO=18,4-11,2=7,2(g)
b)nH2SO4=nH2=0,2(mol)
=>mH2SO4 7%=0,2.98=19,6(g)
=>mH2SO4 =19,6:7%=280(g)
c)mFeSO4=0,2.152=30,4(g)
mdd sau pư=18,4+280-0,2.2=298(g)
=>C%FeSO4=\(\frac{30,4}{298}.100\%\)=10,2%
mHCl=(7,3*300)/100=21,9 g =>nHCl=0,6 mol
mH2SO4=(100*19,6)/100=19,6g =>nH2SO4=0,2 mol
PT HCl+NaOH-> NaCl+ H2O
mol 0,6 0,6 0,6
2NaOH+ H2SO4->Na2SO4+ H2O
mol 0,4 0,2 0,2
m NaOH=(0,4+0,6)*40=40g =>mdd NaOH=(40*100)/5=800g
C%NaCl=(0,6*58,5*100%)/(300+800+100)=2,925%
C%Na2SO4=(0,2*142*100%)/(300+800+100)=2,367%
\(n_{FeCl3}=\dfrac{32,5}{162,5}=0,2\)(mol)
Pt: FeO + 2HCl \(\rightarrow FeCl_2+H_2O\)
\(Fe_3O_4+8HCl\rightarrow FeCl_2+2FeCl_3+4H_2O\)
0,1 0,2 (mol)
\(\Rightarrow\)nFe3O4=0,1(mol)
\(\Rightarrow m_{Fe3O4}=0,1.232=23,2\left(g\right)\)
\(\Rightarrow m_{FeO}=37,6-23,2\)=14,4(g)
\(\Rightarrow\)nFeO = 0,2(mol)
\(\Rightarrow n_{HCl}=\)0,2.2+0,1.8=1,2(mol)
\(\Rightarrow m_{HCl}=43,8\left(g\right)\)
\(\Rightarrow a=\)\(\dfrac{43,8.100}{5}\)=876
Ta có mdd sau PƯ = 37,6+876=913,6
\(\Rightarrow C\%_{FeCl2}=\dfrac{\left(0,2+0,1\right).127}{913,6}\).100%=4,17
\(\Rightarrow C\%_{FeCl3}=\dfrac{0,2.162,5}{913,6}.100\%=3,56\%\)
Bài 1: Mg + HCl -----> MgCl2 + H2
Ag + HCl---/---> ko x/r pứ
a,T/có: nH2=5,6/22,4=0,25 mol
Theo PTHH: n Mg =n H2=0,25 mol
=>m Mg=0,25.24=6g
=>mAg = 27,6 - 6 = 21,6g
=>%m Mg=6.100%/27,6=21,7%
=>%m Ag =100%-21,7% = 78,3%
b.Theo PTHH: nHCl=2n H2 = 0,5 mol
=>m HCl=0,5.36,5=18,25g
=>m dd HCl=18,25.100% / 14,6% =125g
c.Theo PTHH:n MgCl2=nH2=0,25 mol
Theo định luật bt khlg: mdd sau pứ=m hỗn hợp + m dd HCl-m H2
=27,6+125-(0,25.2)=152,1g
=>C% dd MgCl2=0,25.95/152,1.100%=15,6%
d.Chất rắn:Ag
2Ag + 2H2SO4(đ,n)---------> Ag2SO4+SO2+2H2O
T/có: n Ag = 21,6/108=0,2 mol
Theo PTHH: n SO2 = 1/2 n Ag=0,1 mol
=>V H2(ở đktc)=0,1.22,4=2,24l
Theo PTHH: n H2SO4=nAg=0,2 mol
=>mH2SO4=0,2.98=19,6g
=>m ddH2SO4=19,6.100% /80%=24,5g
a)
NaCl + AgNO3 -> NaNO3 + AgCl↓
0,1 0,1 0,1 0,1
b)
mNaCl = \(\frac{100.5,85}{100}=5,85\)=> nNaCl = 5,85 : 58,5 = 0,1 mol
mAgCl = 0,1 . 143,5 = 14,35 g
c)
mNaNO3 = 0,1 . 85 = 8,5 g
a, NaCl + AgNO3 --> NaNO3 +AgCl
1mol 1mol 1mol 1mol
0,1mol 0,1mol 0,1mol 0,1mol
b, mNaCl=100.5,85%=5,85g
nNaCl=\(\frac{5,85}{58,5}\)=0,1(mol)
mAgNO3=0,1.170=17g
mNaNO3=0,1.85=8,5g
mAgCl=0,1.143,5=14,35g
a. Ta có: \(n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)\)
Ta lại có: \(C_{\%_{HCl}}=\dfrac{m_{ct_{HCl}}}{100}.100\%=7,3\%\)
=> mHCl = 7,3(g)
=> \(n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)
PTHH:
Fe3O4 + 8HCl ---> FeCl2 + 2FeCl3 + 4H2O
1 ---> 8
0,1 ---> 0,2
=> \(\dfrac{0,1}{1}>\dfrac{0,2}{8}\)
Vậy Fe3O4 dư
=> mdư = 23,2 - 7,3 = 15,9 (g)
b. Theo PT: \(n_{FeCl_2}=\dfrac{1}{8}.n_{HCl}=\dfrac{1}{8}.0,2=0,025\left(mol\right)\)
=> \(m_{FeCl_2}=0,025.127=3,175\left(g\right)\)
Theo PT: \(n_{FeCl_3}=\dfrac{1}{4}.n_{HCl}=\dfrac{1}{4}.0,2=0,05\left(mol\right)\)
=> \(m_{FeCl_3}=0,05.162,5=8,125\left(g\right)\)
=> \(m_{muối}=8,125+3,175=11,3\left(g\right)\)
c. Ta có: mdung dịch sau PỨ = \(23,2+100=123,2\left(g\right)\)
Theo PT: \(n_{H_2O}=\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)
=> \(m_{H_2O}=0,1.18=1,8\left(g\right)\)
mcác chất sau PỨ = 1,8 + 11,3 = 13,1(g)
=> \(C_{\%_{sauPỨ}}=\dfrac{13,1}{123,2}.100\%=10,63\%\)