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Bài 1:
\(a.n_{NaOH\left(tổng\right)}=0,05.1+0,2.0,2=0,09\left(mol\right)\\ V_{ddNaOH\left(tổng\right)}=50+200=250\left(ml\right)=0,25\left(l\right)\\ C_{MddNaOH\left(cuối\right)}=\dfrac{0,09}{0,25}=0,36\left(M\right)\\ b.n_{HCl}=0,5.0,02=0,01\left(mol\right)\\ n_{H_2SO_4}=0,08.0,2=0,016\left(mol\right)\\ V_{ddsau}=20+80=100\left(ml\right)=0,1\left(l\right)\\ C_{MddH_2SO_4}=\dfrac{0,016}{0,1}=0,16\left(M\right)\\ C_{MddHCl}=\dfrac{0,01}{0,1}=0,1\left(M\right)\)
Bài 2:
\(a.m_{H_2SO_4}=29,4.10\%=2,94\left(g\right)\\ b.n_{H_2SO_4}=\dfrac{2,94}{98}=0,03\left(mol\right)\\ n_{Fe}=\dfrac{0,56}{56}=0,01\left(mol\right)\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ Vì:\dfrac{0,01}{1}< \dfrac{0,03}{1}\Rightarrow H_2SO_4dư\\ n_{H_2SO_4\left(dư\right)}=0,03-0,01=0,02\left(mol\right)\\ m_{H_2SO_4\left(dư\right)}=0,02.98=1,96\left(g\right)\\ n_{H_2}=n_{Fe}=0,01\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,01.22,4=0,224\left(l\right)\)
\(n_{NaOH}=\dfrac{200\cdot4\%}{40}=0.2\left(mol\right)\)
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\)
\(0.2..............0.1..............0.1\)
\(V_{dd_{H_2SO_4}}=\dfrac{0.1}{0.2}=0.5\left(l\right)\)
\(m_{Na_2SO_4}=0.1\cdot142=14.2\left(g\right)\)
\(m_{dd}=200+510=710\left(g\right)\)
\(C\%_{Na_2SO_4}=\dfrac{14.2}{710}\cdot100\%=2\%\)
Ta có: mNaOH = 200.4% = 8 (g)
\(\Rightarrow n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)
PT: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
_____0,2______0,1_______0,1 (mol)
a, \(V_{ddH_2SO_4}=\dfrac{0,1}{0,2}=0,5\left(l\right)\)
b, Chất có trong dd sau pư là Na2SO4.
Ta có: m dd sau pư = m dd NaOH + m dd H2SO4 = 200 + 510 = 710 (g)
\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{0,1.142}{710}.100\%=2\%\)
Bạn tham khảo nhé!
a)Gọi C% của dd sau khi trộn là x (%)
Ta có sơ đồ đường chéo:
50g NaOH 8% ........................... 20-x
................ x (%)............
450g NaOH 20%........................... x-8
=> \(\dfrac{50}{250}=\dfrac{20-x}{x-8}\)<=> x = 18,8 %
b) CM = \(\dfrac{18,8.1,1}{40}\)= 0,517 (mol/lít)
mNaOH = \(\dfrac{\left(45+450\right).18,8}{100}\)100 = 94 (g) => nNaOH = \(\dfrac{94}{40}\) = 2,35(mol)
=> Vdd = \(\dfrac{2,35}{0,517}\)≈ 4,55 (l)
Câu 1:
PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)
Ta có: \(n_{HCl}=0,2\cdot2=0,4\left(mol\right)=n_{NaOH}\)
\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,4}{0,2}=2\left(M\right)\)
Câu 2: Bạn xem lại đề !!
a)
\(n_{H_2}=\dfrac{0,6}{2}=0,3\left(mol\right)\)
PTHH: 2M + 3H2SO4 --> M2(SO4)3 + 3H2
0,2<----0,3<--------0,1<-------0,3
=> \(M_M=\dfrac{5,4}{0,2}=27\left(g/mol\right)\)
=> M là Al
b) \(C\%_{dd.H_2SO_4}=\dfrac{0,3.98}{395,2}.100\%=7,44\%\)
c)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
0,2-->0,6
=> \(V_{dd.HCl}=\dfrac{0,6}{1}=0,6\left(l\right)=600\left(ml\right)\)
\(n_{H_2SO_4}=0,1.3=0,3\left(mol\right)\)
PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
Theo PT: \(n_{Zn}=n_{H_2SO_4}=0,3\left(mol\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Theo PT: \(n_{ZnCl_2}=n_{Zn}=0,3\left(mol\right)\Rightarrow m_{ZnCl_2}=0,3.136=40,8\left(g\right)\)
\(m_{HCl}=\dfrac{300.7,3\%}{100\%}=21,9g\\ n_{HCl}=\dfrac{21,9}{36,5}=0,6mol\\ HCl+NaOH\rightarrow NaCl+H_2O\left(1\right)\\ n_{NaOH\left(1\right)}=n_{HCl}=0,6mol\\ m_{H_2SO_4}=\dfrac{200.9,8\%}{100\%}=19,6g\\ n_{H_2SO_4}=\dfrac{19,6}{98}=0,2mol\\ H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\left(2\right)\\ n_{NaOH\left(2\right)}=0,2.2=0,4mol\\ n_{NaOH}=0,4+0,6=1mol\\ m_{NaOH}=1.40=40g\\ m_{ddNaOH}=\dfrac{40}{5\%}\cdot100\%=800g\)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ a,Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{HCl}=2.0,2=0,4\left(mol\right)\\ b,m_{ddHCl}=\dfrac{0,4.36,5.100}{20}=73\left(g\right)\\ c,n_{H_2}=n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\\ m_{ddsau}=11,2+73-0,2.2=83,8\left(g\right)\\ C\%_{ddFeCl_2}=\dfrac{0,2.127}{83,8}.100\approx30,31\%\)