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a,\(m_{BaCl_2}=208.15\%=31,2\left(g\right)\Rightarrow n_{BaCl_2}=\dfrac{31,2}{208}=0,15\left(mol\right)\)
\(m_{H_2SO_4}=150.19,6\%=29,4\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\left(mol\right)\)
PTHH: BaCl2 + H2SO4 → BaSO4 + 2HCl
Mol: 0,15 0,15 0,15 0,3
Ta có: \(\dfrac{0,15}{1}< \dfrac{0,3}{1}\) ⇒ BaCl2 hết, H2SO4 dư
\(m_{H_2SO_4dư}=\left(0,3-0,15\right).98=14,7\left(g\right)\)
b, \(m_{BaSO_4}=0,15.233=34,95\left(g\right)\)
\(m_{HCl}=0,3.36,5=10,95\left(g\right)\)
PTHH: \(CuSO_4+2NaOH\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\)
Ta có: \(\left\{{}\begin{matrix}n_{CuSO_4}=0,08\cdot3,5=0,28\left(mol\right)\\n_{NaOH}=0,12\cdot1,5=0,18\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,28}{1}>\dfrac{0,18}{2}\) \(\Rightarrow\) CuSO4 còn dư
\(\Rightarrow\left\{{}\begin{matrix}n_{Na_2SO_4}=0,09\left(mol\right)=n_{Cu\left(OH\right)_2}\\n_{CuSO_4\left(dư\right)}=0,19\left(mol\right)\\\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Na_2SO_4}=0,09\cdot142=12,78\left(g\right)\\m_{Cu\left(OH\right)_2}=0,09\cdot98=8,82\left(g\right)\\m_{CuSO_4\left(dư\right)}=0,19\cdot160=30,4\left(g\right)\end{matrix}\right.\)
Mặt khác: \(\left\{{}\begin{matrix}C_{M_{Na_2SO_4}}=\dfrac{0,09}{0,08+0,12}=0,45\left(M\right)\\C_{M_{CuSO_4\left(dư\right)}}=\dfrac{0,19}{0,08+0,12}=0,95\left(M\right)\end{matrix}\right.\)
\(n_{BaCl_2}=\dfrac{31,2}{208}=0,15mol\)
\(BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)
0,15 0,15 0,15 0,3
a)\(m_{BaSO_4}=0,15\cdot233=34,95\left(g\right)\)
b)\(m_{H_2SO_4}=0,15\cdot98=14,7\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{14,7}{19,6}\cdot100=75\left(g\right)\)
c)\(m_{HCl}=0,3\cdot36,5=10,95\left(g\right)\)
\(m_{ddsau}=31,2+75-34,95=71,25\left(g\right)\)
\(\Rightarrow C\%_{HCl}=\dfrac{10,95}{71,25}\cdot100\%=15,37\%\)
\(n_{BaCl_2}=\dfrac{150.16,64\%}{137+35.2}=0,12\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{100.14,7\%}{98}=0,15\left(mol\right)\)
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
\(\Rightarrow n_{NaOH}=0,24+0,06=0,3\left(mol\right)\)
\(V_{NaOH}=0,3.2=0,6\left(l\right)\)
a/ \(n_{KOH}=0,2.1=0,2\left(mol\right);n_{H_2SO_4}=0,3.1=0,3\left(mol\right)\)
PTHH: 2KOH + H2SO4 → K2SO4 + 2H2O
Mol: 0,2 0,1 0,1
Ta có: \(\dfrac{0,2}{2}< \dfrac{0,3}{1}\) ⇒ KOH hết, H2SO4 dư
b/ \(m_{H_2SO_4dư}=\left(0,3-0,1\right).98=19,6\left(g\right)\)
c/ Vdd sau pứ = 0,2 + 0,3 = 0,5 (l)
d/ \(C_{M_{ddK_2SO_4}}=\dfrac{0,1}{0,5}=0,2M\)
\(C_{M_{ddH_2SO_4dư}}=\dfrac{0,3-0,1}{0,5}=0,4M\)
Bài 6 :
a) Pt : \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O|\)
1 1 1 1
a 2a 0,2
\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O|\)
1 3 1 3
b 3b 0,1
b) Gọi a là số mol của MgO
b là số mol của Al2O3
\(m_{MgO}+m_{Al2O3}=18,2\left(g\right)\)
⇒ \(n_{MgO}.M_{MgO}+n_{Al2O3}.M_{Al2O3}=18,2g\)
⇒ 40a + 102b = 18,2g
Ta có : \(m_{ct}=\dfrac{19,6.250}{100}=49\left(g\right)\)
\(n_{H2SO4}=\dfrac{49}{98}=0,5\left(mol\right)\)
⇒ 1a + 3b = 0,5 (2)
Từ (1),(2), ta có hệ phương trình :
40a + 102b = 18,2g
1a + 3b = 0,5
⇒ \(\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)
\(m_{MgO}=0,2.40=8\left(g\right)\)
\(m_{Al2O3}=0,1.102=10,2\left(g\right)\)
d) Có : \(n_{MgO}=0,2\left(mol\right)\Rightarrow n_{MgSO4}=0,2\left(mol\right)\)
\(n_{Al2O3}=0,1\left(mol\right)\Rightarrow n_{Al2\left(SO4\right)3}=0,1\left(mol\right)\)
\(m_{MgSO4}=0,2.120=24\left(g\right)\)
\(m_{Al2\left(SO4\right)3}=0,1.342=34,2\left(g\right)\)
\(m_{ddspu}=18,2+250=268,2\left(g\right)\)
\(C_{MgSO4}=\dfrac{24.100}{268,2}=8,95\)0/0
\(C_{Al2\left(SO4\right)3}=\dfrac{34,2.100}{268,2}=12,75\)0/0
e) \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O|\)
2 1 1 2
1 0,5
\(n_{NaOH}=\dfrac{0,5.2}{1}=1\left(mol\right)\)
\(m_{NaOH}=1.40=40\left(g\right)\)
\(m_{ddnaOH}=\dfrac{40.100}{12}=333,33\left(g\right)\)
\(V_{ddNaOH}=\dfrac{333,33}{1,1}=303,2\left(ml\right)\)
Chúc bạn học tốt
\(n_{BaCl_2}=\dfrac{208.15\%}{208}=0,15\left(mol\right)\\ n_{H_2SO_4}=\dfrac{150.19,6\%}{98}=0,3\left(mol\right)\\ BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\\ Vì:\dfrac{0,15}{1}< \dfrac{0,3}{1}\Rightarrow H_2SO_4dư\\ n_{H_2SO_4\left(p.ứ\right)}=n_{BaSO_4}=n_{BaCl_2}=0,15\left(mol\right)\\ n_{HCl}=2.0,15=0,3\left(mol\right)\\ n_{H_2SO_4\left(dư\right)}=0,13-0,15=0,15\left(mol\right)\\ m_{HCl}=0,3.36,5=10,95\left(g\right)\\ m_{BaSO_4}=233.0,15=34,95\left(g\right)\\ m_{H_2SO_4\left(dư\right)}=0,15.98=14,7\left(g\right)\\ m_{ddsau}=208+150-34,95=323,05\left(g\right)\\ C\%_{ddHCl}=\dfrac{10,95}{323,05}.100\approx3,39\%\)
\(C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{14,7}{323,05}.100\approx4,55\%\)