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Ta có: \(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
____0,4_____0,8____0,4_____0,4 (mol)
a, \(m_{HCl}=0,8.36,5=29,2\left(g\right)\)
b, \(m_{FeCl_2}=0,4.127=50,8\left(g\right)\)
c, \(m_{H_2}=0,4.2=0,8\left(g\right)\)
d, \(m_{ddHCl}=\dfrac{29,2}{14,6\%}=200\left(g\right)\)
m dd sau pư = 22,4 + 200 - 0,8 = 221,6 (g)
e, \(C\%_{FeCl_2}=\dfrac{50,8}{221,6}.100\%\approx22,92\%\)
Ta có \(n_{NaOH}=C_M.V=0,2.1=0,2\left(mol\right)\);
\(n_{H_2SO_4}=C_M.V=0,5.0,3=0,15\left(mol\right)\);
PTHH phản ứng
2NaOH + H2SO4 ---> Na2SO4 + 2H2O
2 : 1 : 2 :1
Nhận thấy \(\dfrac{n_{NaOH}}{2}< \dfrac{n_{H_2SO_4}}{1}\)
=> H2SO4 dư
\(m_{Na_2SO_4}=n.M=0,2.174=34,8\)(g)
b) \(n_{H_2SO_4dư}=0,15-0,1=0,05\) (mol)
=> \(C_{MH_2SO_4}=\dfrac{n}{V}=\dfrac{0,05}{0,5}=0,1\left(M\right)\)
\(C_{MNa_2SO_4}=\dfrac{n}{V}=\dfrac{0,2}{0,5}=0,4\left(M\right)\)
\(n_{H_2SO_4}=0,2.0,25=0,05
mol
\)
a) \(H_2SO_4+2KOH
\rightarrow
K_2SO_4+2H_2O\)
Theo PTHH ta có: \(n_{KOH}=2n_{H_2SO_4}=0,05.2=0,1
mol\)
\(\rightarrow V_{dd
KOH}=\frac{n}{C_M}=\frac{0,1}{0,5}=0,2
\left(l\right)=200\left(ml\right)\)
\(n_{K_2SO_4}=n_{H_2SO_4}=0,05\left(mol\right)\)
\(m_{dd
sau}=250+200=450=0,45\left(l\right)\)
\(\rightarrow C_{M
K_2SO_4}=\frac{0,05}{0,45}=\frac{1}{9}M\)
b)\(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)
\(n_{NaOH}=2.n_{H_2SO_4}=0,1\left(mol\right)\rightarrow m_{NaOH}=0,1.40=4\left(g\right)\)
\(m_{dd
NaOH}=\frac{4.100}{20}=20\left(g\right)\)
\(n_{Na_2SO_4}=0,05\left(mol\right)\rightarrow m_{Na_2SO_4}=0,05.142=7,1\left(g\right)\)
Do D=1ml/g -> mdd H2SO4=250(g)
\(C\%_{Na_2SO_4}=\frac{7,1}{250+20}.100=2,63\%\)
mKOH=28(g)
nKOH=0.5(mol)
PTHH:2KOH+H2SO4->K2SO4+2H2O
a)Theo pthh:nH2SO4=1/2 nKOH->nH2SO4=0.25(mol)
mH2SO4=0.25*98=24.5(g)
C%ddH2SO4=24.5/100*100=24.5%
theo pthh:nK2SO4=nH2SO4->nK2SO4=0.25(mol)
mK2SO4=0.25*(39*2+96)=43.5(g)
c)mdd sau phản ứng:200+100=300(g)
d) C% muối=43.5:300*100=14.5%
2) Na2O + H2SO4 --> Na2SO4 + H2O
nNa2SO4 = 0,02 --> mNa2O = 1,24 gam
Gọi CTHH của oxit KL là A2On.
PT: \(A_2O_n+nH_2SO_4\rightarrow A_2\left(SO_4\right)_n+nH_2O\)
Ta có: \(n_{A_2O_n}=\dfrac{4,8}{2M_A+16n}\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=n.n_{A_2O_n}=\dfrac{4,8n}{2M_A+16n}\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=\dfrac{4,8n}{2M_A+16n}.98=\dfrac{470,4n}{2M_A+16n}\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{\dfrac{470,4n}{2M_A+16n}}{10\%}=\dfrac{4704n}{2M_A+16}\left(g\right)\)
⇒ m dd sau pư = \(4,8+\dfrac{4704n}{2M_A+16}\left(g\right)\)
Theo PT: \(n_{A_2\left(SO_4\right)_n}=n_{A_2O_n}=\dfrac{4,8}{2M_A+16n}\left(mol\right)\)
\(\Rightarrow C\%_{A_2\left(SO_4\right)_n}=\dfrac{\dfrac{4,8.\left(2M_A+96n\right)}{2M_A+16}}{4,8+\dfrac{4704n}{2M_A+16n}}.100\%=12,9\%\)
\(\Rightarrow M_A\approx18,65m\)
Với m = 3, MA = 56 (g/mol) là thỏa mãn.
→ A là Fe.
Ta có: \(n_{Fe_2O_3}=\dfrac{4,8}{160}=0,03\left(mol\right)\)
Theo PT: \(n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,03\left(mol\right)\)
Gọi CTHH của muối P là Fe2(SO4)3.nH2O.
Có: H = 80% ⇒ nP = 0,03.80% = 0,024 (mol)
\(\Rightarrow M_P=\dfrac{13,488}{0,024}=562\left(g/mol\right)\)
\(\Rightarrow400+18n=562\Rightarrow n=9\)
Vậy: CTHH của P là Fe2(SO4)3.9H2O
Ta có :
\(\text{nBaCl2=512x20%/208=32/65(mol)}\)
nBaCl2 dư=nH2SO4=100x9,8%/98=0,1(mol)
=>nBaCl2 phản ứng=32/65-0,1=51/130(mol)
Gọi a là m dd A
\(\text{mx17,4%/174+mx14,2%/142=51/130}\)
=>m=196,15(g)
b)m dd spu=196,15+512-51/130x233=616,74(g)
\(\text{C%KCl=2x0,19615x74,5/616,74=4,74%}\)
\(\text{C%NaCl=2x0,19615x58,5/616,74=3,72% }\)
\(a.n_{H_2SO_4}=\dfrac{196.10\%}{98}=0,2\left(mol\right)\\ ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\\ 0,2.......0,2........0,2.......0,2\left(mol\right)\\ m_{ZnSO_4}=161.0,2=32,2\left(g\right)\\ b.m_{ddsau}=196+0,2.81=212,2\left(g\right)\\ c.C\%_{ddZnSO_4}=\dfrac{32,2}{212,2}.100\approx15,174\%\)
câu b 81 ở đâu ra vậy ạ