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\(n_{CuSO_4}=\dfrac{200.16\%}{160}=0,2\left(mol\right)\)
PTHH :
\(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\)
0,2 0,4 0,2 0,2
\(m_{NaOH}=0,4.40=16\left(g\right)\)
\(m_{ddNaOH}=\dfrac{16.100}{10}=160\left(g\right)\)
\(c,m_{Na_2SO_4}=0,2.142=28,4\left(g\right)\)
\(m_{ddNa_2SO_4}=200+160-\left(0,2.98\right)=340,4\left(g\right)\)
\(C\%_{Na_2SO_4}=\dfrac{28,4}{240,4}.100\%\approx8,34\%\)
\(d,PTHH:\)
\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)
0,2 0,2
\(m_{CuO}=0,2.80=16\left(g\right)\)
a, \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
b, \(m_{CuSO_4}=200.16\%=32\left(g\right)\Rightarrow n_{CuSO_4}=\dfrac{32}{160}=0,2\left(mol\right)\)
\(n_{NaOH}=2n_{CuSO_4}=0,4\left(mol\right)\Rightarrow m_{ddNaOH}=\dfrac{0,4.40}{10\%}=160\left(g\right)\)
c, \(n_{Cu\left(OH\right)_2}=n_{Na_2SO_4}=n_{CuSO_4}=0,2\left(mol\right)\)
\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{0,2.142}{200+160-0,2.98}.100\%\approx8,34\%\)
d, \(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)
\(n_{CuO}=n_{Cu\left(OH\right)_2}=0,2\left(mol\right)\Rightarrow m_{CuO}=0,2.80=16\left(g\right)\)
\(n_{BaCl_2}=\dfrac{150.16,64\%}{137+35.2}=0,12\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{100.14,7\%}{98}=0,15\left(mol\right)\)
Phương trình hóa học :
BaCl2 + H2SO4 -----> BaSO4 + 2HCl
Dễ thấy \(\dfrac{n_{BaCl_2}}{1}< \dfrac{n_{H_2SO_4}}{1}\Rightarrow H_2SO_4\text{ dư }0,15-0,12=0,03\left(mol\right)\)
c) Khối lượng kết tủa :
\(m_{BaSO_4}=0,12.233=27,96\) (g)
Khối lượng chất tan : \(m_{HCl}=0,24.36,5=8,76\left(g\right)\) ;
\(m_{H_2SO_4\left(\text{dư}\right)}=0,03.98=2,94\left(g\right)\)
c) \(C\%_{H_2SO_4}\)= \(\dfrac{2,94}{150+100}.100\%=1,176\%\)
\(C\%_{HCl}=\dfrac{8,76}{150+100}.100\%=3.504\%\)
d) NaOH + HCl ---> NaCl + H2O
0,24 <-- 0,24
mol mol
2NaOH + H2SO4 ---> Na2SO4 + 2H2O
0,06 mol <-- 0,03 mol
\(\Rightarrow n_{NaOH}=0,24+0,06=0,3\left(mol\right)\)
\(V_{NaOH}=0,3.2=0,6\left(l\right)\)
\(n_{FeSO_4}=\dfrac{15,2}{152}=0,1mol\)
\(n_{NaOH}=\dfrac{40}{40}=1mol\)
\(FeSO_4+2NaOH\rightarrow Fe\left(OH\right)_2+Na_2SO_4\)
0,1 < 1 ( mol )
0,1 0,2 0,1 0,1 ( mol )
\(m_{Fe\left(OH\right)_2}=0,1.90=9g\)
\(m_{NaOH\left(dư\right)}=\left(1-0,2\right).40=32g\)
Cách 1.
\(m_{Na_2SO_4}=0,1.142=14,2g\)
Cách 2.
\(m_{NaOH\left(pứ\right)}=0,2.40=8g\)
Áp dụng ĐL BTKL, ta có:
\(m_{FeSO_4}+m_{NaOH}=m_{Fe\left(OH\right)_2}+m_{Na_2SO_4}\)
\(\rightarrow m_{Na_2SO_4}=15,2+8-9=14,2g\)
\(nCaCl_2=\dfrac{2,22}{111}=0,02\left(mol\right)\)
\(nAgNO_3=\dfrac{1,7}{170}=0,01\left(mol\right)\)
\(CaCl_2+2AgNO_3\rightarrow2AgCl+Ca\left(NO_3\right)_2\)
1 2 2 1 (mol)
0,005 0,01 0,01 0,005
LTL : \(\dfrac{0,02}{1}>\dfrac{0,01}{2}\)
=> CaCl2 dư , AgNO3 đủ
\(m_{kt}=mAgCl=0,01.143,5=1,435\left(g\right)\)
c1:
\(m_{\left(muối\right)}=m_{Ca\left(NO_3\right)_2}=0,005.164=0,82\left(g\right)\)
c2:
BTKL:
\(mCaCl_{2\left(đủvspứ\right)}=0,005.111=0,555\left(g\right)\)
\(mCaCl_2+mAgNO_3=mAgCl+mCa\left(NO_3\right)_2\)
0,555 + 1,7 = 1,435 + \(mCa\left(NO_3\right)_2\)
\(\Rightarrow mCa\left(NO_3\right)_2=0,555+1,7-1,435=0,82\left(g\right)\)
a) PTHH: \(KOH+HCl\rightarrow KCl+H_2O\)
b) Ta có: \(n_{KCl}=0,15\cdot0,5=0,075\left(mol\right)=n_{KOH}\) \(\Rightarrow m_{KOH}=0,075\cdot56=4,2\left(g\right)\)
c) PTHH: \(KCl+AgNO_3\rightarrow KNO_3+AgCl\downarrow\)
Theo PTHH: \(n_{KCl}=0,075\left(mol\right)=n_{AgNO_3\left(p.ứ\right)}=n_{KNO_3}=n_{AgCl}\)
\(\Rightarrow n_{AgNO_3\left(dư\right)}=0,075\cdot120\%-0,075=0,015\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{AgCl}=0,075\cdot143,5=10,7625\left(g\right)\\C_{M_{KNO_3}}=\dfrac{0,075}{0,5+2}=0,03\left(M\right)\\C_{M_{AgNO_3\left(dư\right)}}=\dfrac{0,015}{2,5}=0,006\left(M\right)\end{matrix}\right.\)
d) Coi như khi cô cạn không bị hao hụt muối
Ta có: \(m_{muối.khan}=m_{KNO_3}+m_{AgNO_3\left(dư\right)}=0,075\cdot101+0,015\cdot170=10,125\left(g\right)\)
\(n_{CuSO_4}=\dfrac{48}{160}=0,3mol\)
\(n_{BaCl_2}=\dfrac{82,2}{208}=0,39mol\)
\(CuSO_4+BaCl_2\rightarrow BaSO_4+CuCl_2\)
0,3 < 0,39 ( mol )
0,3 0,3 0,3 0,3 ( mol )
\(m_{BaCl_2\left(dư\right)}=\left(0,39-0,3\right).208=18,72g\)
\(m_{BaSO_4}=0,3.233=69,9g\)
Cách 1. \(m_{CuCl_2}=0,3.135=40,5g\)
Cách 2. \(m_{BaCl_2\left(pứ\right)}=0,3.208=62,4g\)
Áp dụng ĐL BTKL, ta có:
\(m_{CuSO_4}+m_{BaCl_2}=m_{BaSO_4}+m_{CuCl_2}\)
\(\rightarrow m_{CuCl_2}=48+62,4-69,9=40,5g\)