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nNa2CO3 = 10,6 / 106 = 0,1 (mol)
Na2CO3 + 2CH3COOH -> 2CH3COONa + H2O + CO2
0,1 0,2 0,2 0,1
mdd CH3COOH = 0,2 * 60 / 5 * 100 = 240 (gam)
CO2 + Ca(OH)2 -> CaCO3 + H2O
0,1 0,1
mCaCO3 = 0,1 * 100 = 10 (gam)
mdd = 240 + 10,6 - 0,1 * 44 = 246,2 (gam)
C% = 82 * 0,2 / 246,2 * 100% = 6,66%
\(n_{Na_2CO_3}=\dfrac{10,6}{106}=0,1\left(mol\right)\)
PTHH:
Na2CO3 + 2CH3COOH ---> 2CH3COONa + CO2 + H2O
0,1---------->0,2----------------->0,2--------------->0,1
CO2 + Ca(OH)2 ---> CaCO3 + H2O
0,1------------------------->0,1
=> \(\left\{{}\begin{matrix}m_{ddCH_3COOH}=\dfrac{0,2.60}{5\%}=240\left(g\right)\\m_{CaCO_3}=0,1.100=10\left(g\right)\end{matrix}\right.\)
\(m_{dd}=10,6+240-0,1.44=246,2\left(g\right)\\ C\%_{CH_3COONa}=\dfrac{82.0,2}{246,2}.100\%=6,66\%\)
Mg + 2HCl → MgCl2 + H2 (1)
Al2O3 + 6HCl → 2AlCl3 + 3H2O (2)
nH2 = 2,8/22,4 = 0,125 mol
Theo tỉ lệ phản ứng (1) => nMg = nH2 = 0,125 mol
<=> mMg = 0,125 .24 = 3 gam và mAl2O3 = 8,1 - 3 =5,1 gam
%mMg = \(\dfrac{3}{8,1}\).100% = 37,03% => %mAl2O3 = 100 - 37,03 = 62,97%
b) nAl2O3 = \(\dfrac{5,1}{102}\)= 0,05 mol
=> nHCl pư = 2nMg + 6nAl2O3 = 0,55 mol
mHCl = 0,55.36,5 = 20,075 gam
=> mdung dịch HCl 18% = \(\dfrac{20,075}{18\%}\)= 111,53 gam
a)
$n_{Al} = 0,3(mol)$
$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
Theo PTHH :
$n_{H_2SO_4} = \dfrac{3}{2}n_{Al} = 0,45(mol)$
$m_{dd\ H_2SO_4} = \dfrac{0,45.98}{12,25\%} = 360(gam)$
b)
$n_{H_2} = n_{H_2SO_4} = 0,45(mol)$
$V_{H_2} = 0,45.22,4 = 10,08(lít)$
c)
$n_{Al_2(SO_4)_3} = 0,15(mol)$
$m_{dd\ sau\ pư} = 8,1 + 360 - 0,45.2 = 367,2(gam)$
$C\%_{Al_2(SO_4)_3} = \dfrac{0,15.342}{367,2}.100\% = 14\%$
Bài 4 :
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,2 0,4 0,2 0,2
b) \(n_{H2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)
c) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)
⇒ \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(C_{ddHCl}=\dfrac{14,6.100}{100}=14,6\)0/0
d) \(n_{ZnCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
⇒ \(m_{ZnCl2}=0,2.136=27,2\left(g\right)\)
\(m_{ddspu}=13+100-\left(0,2.2\right)=112,6\left(g\right)\)
\(C_{ZnCl2}=\dfrac{27,2.100}{112,6}=24,16\)0/0
Chúc bạn học tốt
Bài 3 :
\(n_{Mg}=\dfrac{12}{24}=0,5\left(mol\right)\)
a) Pt : \(Mg+H_2SO_4\rightarrow MgSO_4+H_2|\)
1 1 1 1
0,5 0,5 0,5 0,5
b) \(n_{H2}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,5.22,4=11,2\left(l\right)\)
c) \(n_{H2SO4}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)
⇒ \(m_{H2SO4}=0,5.98=49\left(g\right)\)
\(C_{ddH2SO4}=\dfrac{49.100}{200}=24,5\)0/0
d) \(n_{MgSO4}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)
⇒ \(m_{MgSO4}=0,5.120=60\left(g\right)\)
\(m_{ddspu}=12+200-\left(0,5.2\right)=211\left(g\right)\)
\(C_{MgSO4}=\dfrac{60.100}{211}=28,44\)0/0
Chúc bạn học tốt
Câu 9 :
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
a) Pt : \(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\)
0,2 0,4 0,2 0,2
→ \(V_{H2\left(dtkc\right)}=0,2.22,4=4,48\left(l\right)\)
b) \(V_{ddCH3COOH}=\dfrac{0,4}{2}=0,2\left(l\right)\)
c) \(m_{\left(CH3COO\right)2Mg}=0,2.101=20,2\left(g\right)\)
d) Pt : \(CH_3COOH+KOH\rightarrow CH_3COOK+H_2O\)
0,4 0,4
\(C_{MddKOH}=\dfrac{0,4}{0,2}=2\left(M\right)\)
Chúc bạn học tốt
1)
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
____0,1----->0,15
=> mH2SO4 = 0,15.98 = 14,7(g)
=> \(C\%=\dfrac{14,7}{250}.100\%=5,88\%\)
2)
\(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)
PTHH: Na2CO3 + 2HCl --> 2NaCl + CO2 + H2O
_______0,2------------------------------>0,2
=> VCO2 = 0,2.22,4 = 4,48(l)
3)
\(n_A=\dfrac{18,4}{M_A}\left(mol\right)\)
PTHH: 2A + Cl2 --to--> 2ACl
____\(\dfrac{18,4}{M_A}\)---------->\(\dfrac{18,4}{M_A}\)
=> \(\dfrac{18,4}{M_A}\left(M_A+35,5\right)=46,8=>M_A=23\left(Na\right)\)
4)
nHCl = 0,2.3 = 0,6(mol)
PTHH: M + 2HCl --> MCl2 + H2
____0,3<-----0,6
=> \(M_M=\dfrac{7,2}{0,3}=24\left(Mg\right)\)
a) 2CH3COOH + Na2CO3 --> 2CH3COONa + CO2 + H2O
b) \(n_{CH_3COOH}=\dfrac{200.12\%}{60}=0,4\left(mol\right)\)
PTHH: 2CH3COOH + Na2CO3 --> 2CH3COONa + CO2 + H2O
0,4------->0,2------------------------->0,2
=> \(m_{Na_2CO_3}=0,2.106=21,2\left(g\right)\)
=> \(m_{dd.Na_2CO_3}=\dfrac{21,2.100}{50}=42,4\left(g\right)\)
c) \(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)
a, \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
\(m_{HCl}=109,5.10\%=10,95\left(g\right)\Rightarrow n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)
PTHH: 2Al + 6HCl → 2AlCl3 + 3H2
Mol: 0,1 0,3 0,1 0,15
Ta có: \(\dfrac{0,1}{2}=\dfrac{0,3}{6}\) ⇒ Al hết, HCl hết
\(V_{H_2}=0,15.22,4=3,36\left(l\right)\)
b, \(m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\)
c, mdd sau pứ = 2,7 + 109,5 - 0,15.2 = 111,9 (g)
\(C\%_{ddAlCl_3}=\dfrac{13,35.100\%}{111,9}=11,93\%\)
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\n_{HCl}=\dfrac{109,5\cdot10\%}{36,5}=0,3\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,1}{2}=\dfrac{0,3}{6}\) \(\Rightarrow\) Al và HCl đều p/ứ hết
\(\Rightarrow\left\{{}\begin{matrix}n_{AlCl_3}=0,1\left(mol\right)\\n_{H_2}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{AlCl_3}=0,1\cdot133,5=13,35\left(g\right)\\V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\\m_{H_2}=0,15\cdot2=0,3\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{Al}+m_{ddHCl}-m_{H_2}=111,9\left(g\right)\)
\(\Rightarrow C\%_{AlCl_3}=\dfrac{13,35}{111,9}\cdot100\%\approx11,93\%\)
a) \(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)
PTHH: Na2CO3 + 2CH3COOH --> 2CH3COONa + CO2 + H2O
0,2--------->0,4--------------->0,4------->0,2
=> VCO2 = 0,2.22,4 = 4,48 (l)
b) \(m_{dd.CH_3COOH}=\dfrac{0,4.60}{12\%}=200\left(g\right)\)
c) mdd sau pư = 21,2 + 200 - 0,2.44 = 212,4 (g)
=> \(C\%_{muối}=\dfrac{0,4.82}{212,4}.100\%=15,44\%\)