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a) $n_{H_2SO_4} = 0,1.4,8 = 0,48(mol)$
$Al_2O_3 + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2O$
$n_{Al_2O_3} = \dfrac{1}{3}n_{H_2SO_4} =0,16(mol)$
$m = 0,16.102 = 16,32(gam)$
b)
$n_{Al_2(SO_4)_3} = n_{Al_2O_3} = 0,16(mol)$
$m_{muối} = 0,16.342 = 54,72(gam)$
c)
$Al_2O_3 + 2KOH \to 2KAlO_2 + H_2O$
$n_{KOH} = 2n_{Al_2O_3} = 0,32(mol)$
$V_{dd\ KOH} = \dfrac{0,32}{4,8} = 0,067(lít)$
Ta có: \(n_{H_2SO_4}=0,1.4,8=0,48\left(mol\right)\)
PT: \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
___0,16_____0,48______0,16 (mol)
a, m = mAl2O3 = 0,16.102 = 16,32 (g)
b, mAl2(SO4)3 = 0,16.342 = 54,72 (g)
c, \(Al_2O_3+2KOH\rightarrow2KAlO_2+H_2O\)
____0,16____0,32 (mol)
\(\Rightarrow V_{ddKOH}=\dfrac{0,32}{4,8}=\dfrac{1}{15}\left(l\right)\)
Bạn tham khảo nhé!
a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\) (1)
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\) (2)
\(n_{HCl}=0,45.2=0,9\left(mol\right)\), \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Theo PT (1): \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,2\left(mol\right)\)
\(n_{HCl\left(1\right)}=2n_{H_2}=0,6\left(mol\right)\) \(\Rightarrow n_{HCl\left(2\right)}=0,9-0,6=0,3\left(mol\right)\)
Theo PT: \(n_{Al_2O_3}=\dfrac{1}{6}n_{HCl\left(2\right)}=0,05\left(mol\right)\)
\(\Rightarrow m=m_{Al}+m_{Al_2O_3}=10,5\left(g\right)\)
b, Theo PT (1) + (2): \(n_{AlCl_3}=n_{Al}+2n_{Al_2O_3}=0,3\left(mol\right)\)
Ta có: \(n_{NaOH}=0,5.2=1\left(mol\right)\)
PT: \(AlCl_3+3NaOH\rightarrow Al\left(OH\right)_{3\downarrow}+3NaCl\)
_____0,3_______0,9________0,3 (mol)
\(Al\left(OH\right)_3+NaOH\rightarrow NaAlO_2+2H_2O\)
___0,1________0,1 (mol)
- Hiện tượng: Xuất hiện kết tủa keo trắng, sau đó kết tủa bị hòa tan 1 phần.
\(\Rightarrow n_{Al\left(OH\right)_3}=0,3-0,1=0,2\left(mol\right)\Rightarrow m_{Al\left(OH\right)_3}=0,2.78=15,6\left(g\right)\)
\(NaOH+HCl->NaCl+H_2O\\ 2NaOH+H_2SO_4->Na_2SO_4+2H_2O\\ a.V=\dfrac{0,1.1}{2}=0,05\left(L\right)\\ b.m_{ddH_2SO_4}=\dfrac{0,1.1.98}{2.0,1}=49\left(g\right)\)
nH2=\(\frac{6,72}{22,4}=0,3\)mol
PTHH
M+2HCl--> MCl2+H2
0,3mol<---------------0,3mol
=>MM=\(\frac{19,5}{0,3}=64\)
=> km loại là kẽm (Zn)
b) nNaOH=0,2.1=0,2 mol
PTHH
NaOH+HCl-->NaCl + H2O
0,2 mol--> 0,2 mol
---> thể tích HCl 1M đã dùng là V=\(\frac{0,2+0,3}{1}=0,5\)lít
=> CM(ZnCl2)=\(\frac{0,3}{0,5}=0,6M\)
\(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\)
a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,4 0,8 0,4
b) \(n_{H2}=\dfrac{0,4.1}{1}=0,4\left(mol\right)\)
\(V_{H2\left(dtkc\right)}=0,4.22,4=8,96\left(l\right)\)
c) \(NaOH+HCl\rightarrow NaCl+H_2O|\)
1 1 1 1
0,8 0,8
\(n_{NaOH}=\dfrac{0,8.1}{1}=0,8\left(mol\right)\)
\(V_{ddNaOH}=\dfrac{0,8}{2}=0,4\left(l\right)\)
Chúc bạn học tốt
\(n_{HCl}=0,3.1=0,3\left(mol\right)\\ NaOH+HCl\rightarrow NaCl+H_2O\\ n_{NaOH}=n_{NaCl}=n_{HCl}=0,3\left(mol\right)\\ V_{\text{dd}NaOH}=V=\dfrac{0,3}{2}=0,15\left(l\right)\\ C_{M\text{dd}A}=C_{M\text{dd}NaCl}=\dfrac{0,3}{0,15+0,3}=\dfrac{2}{3}\left(M\right)\)
\(n_{CO_2} = 0,1(mol)\)
CaCO3.MgCO3 + 4HCl → CaCl2 + MgCl2 + 2CO2 + 2H2O(1)
........0,05...............0,2.......................0,05.......0,1..........................(mol)
\(n_{NaOH}= 0,12(mol)\)
MgCl2 + 2NaOH → Mg(OH)2 + 2NaCl(2)
..0,1...........0,2.................................................(mol)
HCl + NaOH → NaCl + H2O(3)
0,02.....0,02................................(mol)
Theo PTHH (1)(3) suy ra :
\(n_{HCl} = 0,2 + 0,02 = 0,22(mol)\\ \Rightarrow V_{dd\ HCl} = \dfrac{0,22}{1} = 0,22(lít)\)
a)
$n_{Al_2O_3} = \dfrac{5,1}{102} = 0,05(mol)$
$Al_2O_3 + 6HCl \to 2AlCl_3 + 3H_2O$
$n_{HCl} = 6n_{Al_2O_3} = 0,3(mol)$
$\Rightarrow V = \dfrac{0,3}{4} = 0,075(lít)$
b)
$Al_2O_3 + 2NaOH \to 2NaAlO_2 + 2H_2O$
$n_{NaOH} = 2n_{Al_2O_3} = 0,1(mol)$
$V_{dd\ NaOH} = \dfrac{0,1}{10} = 0,01(lít)$