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a,\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: CaCO3 + 2HCl → CaCl2 + CO2 + H2O
Mol: 0,1 0,2 0,1
\(m_{CaCO_3}=0,1.100=10\left(g\right)\)
b,\(C\%_{ddHCl}=\dfrac{0,2.36,5.100\%}{150}=4,87\%\)
c,mdd sau pứ= 10+150-0,1.44 = 151,2 (g)
\(C\%_{ddCaCl_2}=\dfrac{0,1.111.100\%}{151,2}=7,34\%\)
\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ 0,05.........0,1..........0,05..........0,05\left(mol\right)\\ a.C\%_{ddHCl}=\dfrac{0,1.36,5}{200}.100=1,825\%\\ b.m_{Zn}=0,05.65=3,25\left(g\right)\\ c.C\%_{ddZnCl_2}=\dfrac{136.0,05}{3,25+200-0,05.2}.100\approx3,347\%\)
Bài 1: a) \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
Theo PT: \(n_{HCl}=6n_{Fe_2O_3}=0,6\left(mol\right)\)
=> \(m_{ddHCl}=\dfrac{0,6.36,5}{14,6\%}=150\left(g\right)\)
b) \(n_{FeCl_3}=2n_{Fe_2O_3}=0,2\left(mol\right)\)
=> \(m_{FeCl_3}=0,2.162,5=32,5\left(g\right)\)
a) \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
Theo PT: \(n_{H_2}=n_{H_2SO_4}=\dfrac{4,9\%.100}{98}=0,05\left(mol\right)\)
=> \(V_{H_2}=0,05.22,4=1,12\left(l\right)\)
b)Theo PT: \(n_{Mg}=n_{H_2SO_4}=0,05\left(mol\right)\)
=> \(m_{Mg}=0,05.24=1,2\left(g\right)\)
\(a,n_{Fe}=\dfrac{11,2}{56}=0,2(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{HCl}=0,4(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,4.36,5}{14,6\%}=100(g)\\ b,n_{H_2}=0,2(mol)\\ \Rightarrow V_{H_2}=0,2.22,4=4,48(l)\\ c,n_{FeCl_2}=0,2(mol)\\ \Rightarrow C\%_{FeCl_2}=\dfrac{0,2.127}{11,2+100-0,2.2}.100\%\approx 22,93\%\)
\(a.n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ n_{CuCl_2}=n_{CuO}=0,1\left(mol\right)\\ n_{HCl}=2.0,1=0,2\left(mol\right)\\ m_{CuCl_2}=135.0,1=13,5\left(g\right)\\ b.m_{HCl}=0,2.36,5=7,3\left(g\right)\\ c.C_{MddHCl}=\dfrac{0,2}{0,2}=1\left(M\right)\)
Em chưa biết làm dạng này như nào em? Vì dạng này rất cơ bản em ạ!
\(m_{ct}=\dfrac{5.200}{100}=10\left(g\right)\)
\(n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\)
Pt : \(NaOH+HCl\rightarrow NaCl+H_2O|\)
1 1 1 1
0,25 0,25 0,25
a) \(n_{HCl}=\dfrac{0,25.1}{1}=0,25\left(mol\right)\)
\(m_{HCl}=0,25.36,5=9,125\left(g\right)\)
\(m_{ddHCl}=\dfrac{9,125.100}{3,65}=250\left(g\right)\)
b) \(n_{NaCl}=\dfrac{0,25.1}{1}=0,25\left(mol\right)\)
⇒ \(m_{NaCl}=0,25.58,5=14,625\left(g\right)\)
\(m_{ddspu}=200+250=450\left(g\right)\)
\(C_{NaCl}=\dfrac{14,625.100}{450}=3,25\)0/0
Chúc bạn học tốt
a) \(n_{NaOH}=\dfrac{200.5\%}{40}=0,25\left(mol\right)\)
PTHH: NaOH + HCl → NaCl + H2O
Mol: 0,25 0,25 0,25
\(m_{ddHCl}=\dfrac{0,25.36,5.100}{3,65}=250\left(g\right)\)
b) mdd sau pứ = 200 + 250 = 450 (g)
\(C\%_{ddNaCl}=\dfrac{0,25.58,5.100\%}{450}=3,25\%\)
PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)
a) Ta có: \(n_{CaCO_3}=\dfrac{2,5}{100}=0,025\left(mol\right)\)
\(\Rightarrow n_{HCl}=0,05mol\) \(\Rightarrow m_{ddHCl}=\dfrac{0,05\cdot36,5}{18\%}\approx10,14\left(g\right)\)
b) Theo PTHH: \(n_{CaCl_2}=n_{CO_2}=n_{CaCO_3}=0,025\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{CaCl_2}=0,025\cdot111=2,775\left(g\right)\\m_{CO_2}=0,025\cdot44=1,1\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd\left(saup/ứ\right)}=m_{Zn}+m_{ddHCl}-m_{CO_2}=11,54\left(g\right)\)
\(\Rightarrow C\%_{CaCl_2}=\dfrac{2,775}{11,54}\cdot100\%\approx24,05\%\)
CaCO3+2HCl→CaCl2+CO2↑ +H2O
\(+n_{CaCO_3}=\dfrac{2,5}{100}=0,025\left(mol\right)\)
\(+n_{HCl}=2n_{CaCO_3}=0,05\left(mol\right)\)
\(+m_{HCl}=0,05.98=4,9\left(gam\right)\)
\(+m_{dungdịchHCl}=\dfrac{4,9}{18}.100\%=27,2\left(gam\right)\)
\(+n_{CaCl}=n_{CaCO_3}=0,025\left(mol\right)\)
\(+m_{CaCl_2}=0,025.111=2,775\left(gam\right)\)
Theo ĐLBTKL ta có:
\(m_{CaCl_2}=2,5+27,2-0,025.44-0,025.18=28,15\left(gam\right)\)
C%=\(\dfrac{2,775}{28,15}.100\%\approx9,85\%\)