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a) 2NaOH + H2SO4 --> Na2SO4 + 2H2O
b) \(m_{NaOH}=\dfrac{200.8}{100}=16\left(g\right)\)
=> \(n_{NaOH}=\dfrac{16}{40}=0,4\left(mol\right)\)
PTHH: 2NaOH + H2SO4 --> Na2SO4 + 2H2O
0,4--->0,2--------->0,2
=> \(m_{Na_2SO_4}=0,2.142=28,4\left(g\right)\)
c) \(m_{H_2SO_4}=0,2.98=19,6\left(g\right)\)
=> \(m_{dd.H_2SO_4}=\dfrac{19,6.100}{9,8}=200\left(g\right)\)
mNaOH = 8% . 200 = 16 (g)
nNaOH = 16/40 = 0,4 (mol)
PTHH: 2NaOH + H2SO4 -> Na2SO4 + 2H2O
Mol: 0,4 ---> 0,2 ---> 0,2 ---> 0,4
mNa2SO4 = 0,2 . 119 = 23,8 (g)
mH2SO4 = 0,2 . 98 = 19,6 (g)
mddH2SO4 = 19,6/9,8% = 200 (g)
Bài 2
Ta có:
nFe=0,2 mol nHCl=0,6 mol
Fe+2HCl=FeCl2+H2
0,2->0,4--->0,2
suy ra sau phản ứng có: 0,2molFeCl2 và 0,2mol HCl dư
CM muối=0,2/0,2=1M
CM axit dư=0,2/0,2=1M
\(n_{H_2SO_4}=C_M.V=0,2.1=0,2\left(mol\right)\)
\(PTHH:H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)
(mol) 1 2 1 2
(mol) 0,2 0,4 0,2 0,4
\(a.m_{NaOH}=n.M=0,4.40=16\left(g\right)\\ \rightarrow m_{ddNaOH}=\frac{16.100}{20}=80\left(g\right)\)
\(b.\\ PTHH:2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
(mol) 2 1 1 2
(mol) 0,4 0,2 0,2 0,4
\(m_{KOH}=n.M=0,4.56=22,4\left(g\right)\\ \rightarrow m_{ddKOH}=\frac{22,4.100}{5,6}=400\left(g\right)\\ \rightarrow V_{ddKOH}=\frac{m}{D}=\frac{400}{1,045}=382,77\left(ml\right)=0,382\left(l\right)\)
\(m_{HCl}=\dfrac{300.7,3\%}{100\%}=21,9g\\ n_{HCl}=\dfrac{21,9}{36,5}=0,6mol\\ HCl+NaOH\rightarrow NaCl+H_2O\left(1\right)\\ n_{NaOH\left(1\right)}=n_{HCl}=0,6mol\\ m_{H_2SO_4}=\dfrac{200.9,8\%}{100\%}=19,6g\\ n_{H_2SO_4}=\dfrac{19,6}{98}=0,2mol\\ H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\left(2\right)\\ n_{NaOH\left(2\right)}=0,2.2=0,4mol\\ n_{NaOH}=0,4+0,6=1mol\\ m_{NaOH}=1.40=40g\\ m_{ddNaOH}=\dfrac{40}{5\%}\cdot100\%=800g\)
2NaOH+ H2SO4 \(\rightarrow\) Na2SO4 + 2H2O (1)
a, nNaOH= CM.V=0,1.0,2=0,02 mol
Theo pt (1) \(n_{H_2SO_4}\)=0,5nNaOH=0,5.0,02=0,01 mol
=> \(m_{H_2SO_4}=\)0,01.98=0,98g
=>\(m_{dd}\)\(_{H_2SO_4}\)\(_{10\%}\)=0,98:10%=9,8g
b, Theo pt n\(_{Na_2SO_4}\)= 0,5.nNaOH=0,01 mol
=> m\(_{Na_2SO_4}\)=0,01.142=1,42g
2NaOH + H2SO4 -> Na2SO4 + 2H2O
nNaOH=0,02(mol)
Theo PTHH ta có:
nNa2SO4=nH2SO4=\(\dfrac{1}{2}\)nNaOH=0,01(mol)
mdd HCl=\(\dfrac{0,01.98}{10\%}=9,8\left(g\right)\)
mNa2SO4=142.0,01=1,42(g)
a, \(n_{NaOH}=0,2.1=0,2\left(mol\right)\)
\(m_{NaOH}=0,2.40=8\left(g\right)\)
b, \(n_{H_2SO_4}=2.0,1=0,2\left(mol\right)\)
\(c,C\%=\dfrac{6}{200}.100\%=3\%\)
\(m_{NaCl}=\dfrac{200.8}{100}=16\left(g\right)\)
a. PTPỨ: H2SO4 + 2NaOH \(\rightarrow\) 2H2O + Na2SO4
b. Ta có : nH2SO4 = \(\frac{1.20}{1000}\) = 0,02 mol
c. Theo phương trình: nNaOH = 2.nH2SO4 = 2.0,02 = 0,04 mol
\(\Rightarrow\) mNaOH = 0,04. 40 = 1,6(g)
d. mdd NaOH = \(\frac{1,6.100}{20}\) = 8(g)
e1. PTHH: H2SO4 + 2KOH \(\rightarrow\) K2SO4 + 2H2O
Ta có: nKOH = 2. nH2SO4 = 2. 0,02 = 0,04 mol
\(\Rightarrow\) mKOH = 0,04.56=2,24(g)
e2. mdd KOH = \(\frac{2,24.100}{5,6}\) = 40(g)
e3. Vdd KOH = \(\frac{40}{1,045}\) \(\approx\) 38,278 ml
Bài 1:
nAl2O3 = \(\dfrac{10,2}{102}=0,1\left(mol\right)\)
Pt: Al2O3 + 6HCl --> 2AlCl3 + 3H2O
0,1 mol---> 0,6 mol-> 0,2 mol
C% dd HCl = \(\dfrac{0,6\times36,5}{182,5}.100\%=12\%\)
mAlCl3 = 0,2 . 133,5 = 26,7 (g)
c) mHCl = \(\dfrac{20\times182,5}{100}=36,5\left(g\right)\)
nHCl = 1 mol
Pt: Al2O3 + 6HCl --> 2AlCl3 + 3H2O
0,1 mol----------------> 0,2 mol
Xét tỉ lệ mol giữa Al2O3 và HCl:
\(\dfrac{0,1}{1}< \dfrac{1}{6}\)
mAlCl3 = 0,2 . 133,5 = 26,7 (g)
Ta có: m dd H2SO4 = 200.1,14 = 228 (g)
\(\Rightarrow m_{H_2SO_4}=228.9,8\%=22,344\left(g\right)\)
\(\Rightarrow n_{H_2SO_4}=\dfrac{22,344}{98}=0,228\left(mol\right)\)
PT: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
a, Theo PT: \(n_{NaOH}=2n_{H_2SO_4}=0,456\left(mol\right)\)
\(\Rightarrow m_{ddNaOH}=\dfrac{0,456.40}{4\%}=456\left(g\right)\)
b, Theo PT: \(n_{Na_2SO_4}=n_{H_2SO_4}=0,228\left(mol\right)\)
\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{0,228.142}{456+228}.100\%\approx4,73\%\)