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PTHH: \(NaCl+AgNO_3\rightarrow NaNO_3+AgCl\downarrow\)
Ta có: \(n_{NaCl}=0,2\cdot0,5=0,1\left(mol\right)=n_{AgNO_3}=n_{AgCl}\)
\(\Rightarrow\left\{{}\begin{matrix}C_{M_{AgNO_3}}=\dfrac{0,1}{0,4}=0,25\left(M\right)\\m_{AgCl}=0,1\cdot143,5=14,35\left(g\right)\end{matrix}\right.\)
a)PTHH: AgNO3 + HCl → AgCl↓ + HNO3
nHCl = 0,2.0,5 = 0,1 mol
=> nAgCl = 0,1 mol = nAgNO3 = 0,1 mol = nHCl phản ứng
<=> mAgCl = 0,1.143,5 = 14,35 gam
mAgNO3 = 0,1.170 = 17 gam
=> mdd AgNO3 = \(\dfrac{17}{6,8\%}\)= 250 gam
b) X + 2HCl --> XCl2 + H2
1,2 gam X tác dụng vừa đủ với 0,1 mol HCl
=> Số mol của 1,2 gam X = 0,05 mol
<=> Mx = \(\dfrac{1,2}{0,05}\)= 24 (g/mol) => X là magie ( Mg )
PTHH: \(Na_2SO_4+CaCl_2\rightarrow2NaCl+CaSO_4\downarrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Na_2SO_4}=0,1\cdot0,5=0,05\left(mol\right)\\n_{CaCl_2}=0,1\cdot0,4=0,04\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Na2SO4 dư
\(\Rightarrow\left\{{}\begin{matrix}n_{CaSO_4}=0,04\left(mol\right)\\n_{NaCl}=0,08\left(mol\right)\\n_{Na_2SO_4\left(dư\right)}=0,01\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaSO_4}=0,04\cdot136=5,44\left(g\right)\\C_{M_{NaCl}}=\dfrac{0,08}{0,1+0,1}=0,4\left(M\right)\\C_{M_{Na_2SO_4\left(dư\right)}}=\dfrac{0,01}{0,2}=0,05\left(M\right)\end{matrix}\right.\)
PTHH: \(2AgNO_3+CaCl_2\rightarrow Ca\left(NO_3\right)_2+2AgCl\downarrow\)
Ta có: \(n_{AgNO_3}=\dfrac{1,7}{170}=0,01\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{AgCl}=0,01\left(mol\right)\\n_{CaCl_2}=n_{Ca\left(NO_3\right)_2}=0,005\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaCl_2}=0,005\cdot111=0,555\left(g\right)\\m_{AgCl}=0,01\cdot143,5=1,435\left(g\right)\\C_{M_{Ca\left(NO_3\right)_2}}=\dfrac{0,005}{0,07+0,03}=0,05\left(M\right)\end{matrix}\right.\)
\(n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)
a) Pt : \(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl|\)
1 2 1 2
0,25 0,5 0,25
b) \(n_{Fe\left(OH\right)2}=\dfrac{0,5.1}{2}=0,25\left(mol\right)\)
⇒ \(m_{Fe\left(OH\right)2}=0,25.90=22,5\left(g\right)\)
c) \(n_{FeCl2}=\dfrac{0,5.1}{2}=0,25\left(mol\right)\)
100ml = 0,1l
\(C_{M_{ddFeCl2}}=\dfrac{0,25}{0,1}=2,5\left(M\right)\)
d) Pt : \(Fe\left(OH\right)_2\underrightarrow{t^o}FeO+H_2O|\)
1 1 1
0,25 0,25
\(n_{FeO}=\dfrac{0,25.1}{1}=0,25\left(mol\right)\)
⇒ \(m_{FeO}=0,25.72=18\left(g\right)\)
Chúc bạn học tốt
nCuCl2 = \(\dfrac{270.15\%}{100\%.135}\)= 0,3(mol)
CuCl2 + 2KOH ➝ Cu(OH)2↓ + 2KCl
0,3 ➝ 0,6 ➝ 0,3 ➝ 0,6 (mol)
a, mCu(OH)2 = 0,3.98= 29,4(g)
b, m dd KOH = \(\dfrac{0,6.56.100\%}{20\%}\)= 168(g)
c, mKCl = 0,6.74,5 = 44,7(g)(*)
Áp dụng định luật bảo toàn khối lượng:
=> mdd KCl= mCuCl2 + m dd KOH - mCu(OH)2
⇔ mdd KCl = 0,3.135+ 168 - 29,4 = 179,1(g)(**)
Từ (*) và (**) ⇒ C%KCl = \(\dfrac{44,7}{179,1}\).100%\(\approx\) 24,96%
PTHH: \(CuCl_2+2KOH\rightarrow2KCl+Cu\left(OH\right)_2\downarrow\)
a) Ta có: \(n_{CuCl_2}=\dfrac{270\cdot15\%}{135}=0,3\left(mol\right)=n_{Cu\left(OH\right)_2}\) \(\Rightarrow m_{Cu\left(OH\right)_2}=0,3\cdot98=29,4\left(g\right)\)
b) Theo PTHH: \(n_{KOH}=2n_{CuCl_2}=0,6mol\) \(\Rightarrow m_{ddKOH}=\dfrac{0,6\cdot57}{20\%}=171\left(g\right)\)
c) Theo PTHH: \(n_{KCl}=n_{KOH}=0,6mol\) \(\Rightarrow m_{KCl}=0,6\cdot74,5=44,7\left(g\right)\)
Mặt khác: \(m_{dd}=m_{ddCuCl_2}+m_{ddKOH}-m_{Cu\left(OH\right)_2}=411,6\left(g\right)\)
\(\Rightarrow C\%_{KCl}=\dfrac{44,7}{411,6}\cdot100\%\approx10,86\%\)
AlCl3 + 3AgNO3 -> 3AgCl + Al(NO3)3 (1)
nAlcl3=m/M= 13.35/133.5=0.1 mol
theo pt
nAgcl= nAgno3=3nAlcl3=3.0.1=0.3 mol; nAl(no3)3=nAlcl3=0.1 mol
100ml = 0.1 lít
=> CM Agno3 = n/V =0.3/0.1=3 M
mAgcl=n.M=0.3.143.5=43.05 g
mAl(no3)3=n.M=0.1.213=21.3 g
nAlCl3=\(\dfrac{13,35}{133,5}=0,1\left(mol\right)\)
pthh:
AlCl3 + 3AgNO3 \(\rightarrow\) Al(NO3)3 + 3AgCl
0,1... .... ....0,3... ... ... ...0,1... ... ... ...0,3 (mol)
a, mAgCl=n.M=0,3.143,5=43,05(g)
b, CMAgNO3=\(\dfrac{n}{V}=\dfrac{0,3}{0,1}=3\left(M\right)\)
c, mmuối Al(NO3)3=n.M=0,1.213=21,3(g)