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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
=> \(M_M=\dfrac{7,2}{0,3}=24\left(Mg\right)\)
PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
\(H_2+\dfrac{1}{2}O_2\xrightarrow[]{t^o}H_2O\)
Ta có: \(n_{H_2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=0,04\left(mol\right)=n_{H_2SO_4}\\n_{O_2}=0,02\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,04\cdot56=2,24\left(g\right)\\C_{M_{H_2SO_4}}=\dfrac{0,04}{0,5}=0,08\left(M\right)\\V_{O_2}=0,02\cdot22,4=0,448\left(l\right)\end{matrix}\right.\)
\(a,n_{\left(CH_3COO\right)_2Mg}=\dfrac{7,1}{142}=0,05\left(mol\right)\)
PTHH: \(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\uparrow\)
0,1<----------------0,05-------------->0,05
\(\rightarrow V_{H_2}=0,05.22,4=1,12\left(l\right)\\ b,C\%_{CH_3COOH}=\dfrac{0,1.60}{100}.100\%=6\%\)
\(c,n_{C_2H_5OH}=\dfrac{6,9}{46}=0,15\left(mol\right)\)
PTHH: \(CH_3COOH+C_2H_5OH\xrightarrow[H_2SO_{4\left(đặc\right)}]{t^o}CH_3COOC_2H_5+H_2O\)
bđ 0,1 0,15
pư 0,1 0,1
spư 0 0,05 0,1
\(\rightarrow m_{este}=0,1.80\%.88=7,04\left(g\right)\)
nCH3COOH=48/60=0,8 mol
2CH3COOH + Fe --> (CH3COO)2Fe + H2
0,8 0,4 0,4 mol
=>m(CH3COO)2Fe=0,4*174=69,6 g
2H2 +O2 --> 2H2O
0,4 0,2 mol
=>VO2=0,2*22,4=4,48 lít
=> V không khí =4,48*5=22,4 lít
PT: \(2CH_3COOH+Fe\rightarrow\left(CH_3COO\right)_2Fe+H_2\)
a, Ta có: \(n_{CH_3COOH}=\dfrac{4,8}{60}=0,08\left(mol\right)\)
Theo PT: \(n_{\left(CH_3COO\right)_2Fe}=\dfrac{1}{2}n_{CH_3COOH}=0,04\left(mol\right)\)
\(\Rightarrow m_{\left(CH_3COO\right)_2Fe}=0,04.174=6,96\left(g\right)\)
b, Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{CH_3COOH}=0,04\left(mol\right)\)
PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)
___0,04__0,02 (mol)
\(\Rightarrow V_{O_2}=0,02.22,4=0,448\left(l\right)\)
\(\Rightarrow V_{kk}=0,448.5=2,24\left(l\right)\)
Bạn tham khảo nhé!
\(n_{Zn}=\dfrac{3,25}{65}=0,05\left(mol\right)\)
\(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\)
0,05 0,1 0,05 0,05 ( mol )
\(V_{H_2}=0,05.22,4=1,12\left(l\right)\)
\(m_{dd_{CH_3COOH}}=\dfrac{0,1.60.100}{20}=30\left(g\right)\)
\(m_{ddspứ}=3,25+30-0,05.2=33,15\left(g\right)\)
\(C\%_{\left(CH_3COO\right)_2Zn}=\dfrac{0,05.183}{33,15}.100=27,6\%\)
2CH3COOH+Mg->(CH3COO)2Mg+H2
0,02---------------0,01-------0,01----------0,01
n muối=0,01mol
=>CM=\(\dfrac{0,02}{0,04}=0,5M\)
=>VH2=0,01.22,4=0,224l
CH3COOH+NaOH->CH3COONa+H2O
0,02--------------0,02
=>VNaOH=\(\dfrac{0,02}{0,75}=0,03l\)
a) \(n_{\left(CH_3COO\right)_2Mg}=\dfrac{1,42}{142}=0,01\left(mol\right)\)
PTHH: Mg + 2CH3COOH --> (CH3COO)2Mg + H2
0,01<-------0,02<------------0,01------->0,01
=> \(C_{M\left(dd.CH_3COOH\right)}=\dfrac{0,02}{0,04}=0,5M\)
b) VH2 = 0,01.22,4 = 0,224 (l)
c)
PTHH: NaOH + CH3COOH --> CH3COONa + H2O
0,02<------0,02
=> \(V_{dd.NaOH}=\dfrac{0,02}{0,75}=\dfrac{2}{75}\left(l\right)=\dfrac{80}{3}\left(ml\right)\)
Ta có: \(n_{Na_2SO_3}=\dfrac{12,6}{126}=0,1\left(mol\right)\)
a. PTHH: Na2SO3 + 2HCl ---> 2NaCl + SO2 + H2O
Theo PT: \(n_{SO_2}=n_{Na_2CO_3}=0,1\left(mol\right)\)
=> \(V_{SO_2}=0,1.22,4=2,24\left(lít\right)\)
b. Theo PT: \(n_{HCl}=2.n_{SO_2}=2.0,1=0,2\left(mol\right)\)
=> \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
=> \(C_{\%_{HCl}}=\dfrac{7,3}{150}.100\%=4,87\%\)
c. Ta có: \(m_{dd_{NaCl}}=n_{Na_2SO_{3_{PỨ}}}=50\left(g\right)\)
Theo PT: \(n_{NaCl}=n_{HCl}=0,2\left(mol\right)\)
=> \(m_{NaCl}=0,2.58,5=11,7\left(g\right)\)
=> \(C_{\%_{NaCl}}=\dfrac{11,7}{50}.100\%=23,4\%\)
Bài 6 :
\(a) Mg + 2CH_3COOH \to (CH_3COO)_2Mg + H_2\\ n_{H_2} = n_{(CH_3COO)_2Mg} = n_{Mg} = \dfrac{9,6}{24} = 0,4(mol)\\ m_{dd\ sau\ pư} = 9,6 + 200 - 0,4.2 = 208,8(gam)\\ C\%_{(CH_3COO)_2Mg} = \dfrac{0,4.142}{208,8}.100\% = 27,2\%\\ b) V_{H_2} = 0,4.22,4 = 8,96(lít)\)
Bài 7 :
\(a) n_{C_2H_5OH} = \dfrac{4,6}{46} = 0,1(mol)\\ C_2H_5OH + 3O_2 \xrightarrow{t^o} 2CO_2 + 3H_2O\\ n_{CO_2} = 2n_{C_2H_5OH} = 0,2(mol)\\ V_{CO_2} = 0,2.22,4 = 4,48(lít)\\ b) n_{O_2} = 3n_{C_2H_5OH} = 0,3(mol)\\ V_{kk} = \dfrac{0,3.22,4}{20\%} = 33,6(lít)\)
Bài 8 :
\(n_{CaCO_3} = \dfrac{12}{100} = 0,12(mol)\\ CaCO_3 + 2CH_3COOH \to (CH_3COO)_2Ca + CO_2 + H_2\\ n_{CH_3COOH} = 2n_{CaCO_3} = 0,24(mol)\\ C\%_{CH_3COOH} = \dfrac{0,24.60}{200}.100\% = 7,2\%\\ b) n_{CO_2} = n_{CaCO_3} = 0,12(mol)\\ V_{CO_2} = 0,12.22,4 = 2,688(lít)\)