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Ta có: \(n_{MgO}=\dfrac{20}{40}=0,5\left(mol\right)\)
PT: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)
_____0,5______1______0,5 (mol)
a, \(m_{HCl}=1.36,5=36,5\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{36,5}{10\%}=365\left(g\right)\)
b, Ta có: m dd sau pư = 20 + 365 = 385 (g)
\(\Rightarrow C\%_{MgCl_2}=\dfrac{0,5.95}{385}.100\%\approx12,34\%\)
\(n_{MgO}=\dfrac{20}{80}=0,25\left(mol\right)\)
PTHH :
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
0,25 0,5 0,25 0,25
\(a,m_{ddHCl}\dfrac{\left(0,5.36,5\right).100}{10}=182,5\left(g\right)\)
\(b,m_{MgCl_2}=0,25.95=23,75\left(g\right)\)
\(m_{ddMgCl_2}=20+182,5=202,5\left(g\right)\)
\(C\%_{MgCl_2}=\dfrac{23,75}{202,5}.100\%=11,73\left(\%\right)\)
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1)\(n_{NaOH}:\dfrac{60.10\%}{100\%.40}=0,15\left(mol\right)\)
KL dung dịch sau p/ư: 60+40=100(g)
\(n_{NaCl}:\dfrac{100.5,85\%}{100\%.58,5}=0,1\left(mol\right)\)
\(NaOH+HCl\rightarrow NaCl+H_2O\)
1...................1...............1.................(mol)
0,1................0,1............0,1...............(mol)
-> NaOH dư
C% dd HCl: \(\dfrac{0,1.36,5}{40}.100\%=9,125\%\)
a) $n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)$
$Fe + 2HCl \to FeCl_2 + H_2$
$n_{HCl} =2 n_{Fe} = 0,2.2 = 0,4(mol)$
$C\%_{HCl} = \dfrac{0,4.36,5}{200}.100\% = 7,3\%$
b) $n_{H_2} = n_{FeCl_2} = n_{Fe} = 0,2(mol)
Sau phản ứng, $m_{dd} = 11,2 + 200 - 0,2.2 = 210,8(gam)$
$C\%_{FeCl_2} = \dfrac{0,2.127}{210,8}.100\% = 12,05\%$
Ta có: \(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)
PT: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
______0,2_____0,4_____0,2 (mol)
a, \(m_{CuCl_2}=0,2.135=27\left(g\right)\)
b, \(m_{HCl}=0,4.36,5=14,6\left(g\right)\Rightarrow C\%_{HCl}=\dfrac{14,6}{300}.100\%\approx4,867\%\)
c, Ta có: m dd sau pư = 16 + 300 = 316 (g)
\(\Rightarrow C\%_{CuCl_2}=\dfrac{27}{316}.100\%\approx8,54\%\)
\(a,n_{Fe}=\dfrac{11,2}{56}=0,2(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{HCl}=0,4(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,4.36,5}{14,6\%}=100(g)\\ b,n_{H_2}=0,2(mol)\\ \Rightarrow V_{H_2}=0,2.22,4=4,48(l)\\ c,n_{FeCl_2}=0,2(mol)\\ \Rightarrow C\%_{FeCl_2}=\dfrac{0,2.127}{11,2+100-0,2.2}.100\%\approx 22,93\%\)
BTKL: mD + mNaHCO3 = mCO2 + mE
mD + 179,88 = 44.0,2 + 492 => mD = 320,92
BTKL: mMg + mddHCl = mH2 + mD
=> 24 . 0,4 + mddHCl = 2 . 0,4 + 320,92 => mddHCl = 312,12
=> C%HCl = 11,69%
m HCl = 480 . 5%=24(g)
tổng m HCl= 24+20=44(g)
m dd sau = 480+ 20= 500 (g)
==>C%(HCl)=\(\frac{44}{500}\) . 100%=8,8%
TH2: mol H2 vẫn là a mol nhé
-----3KOH+AlCl3---->Al(OH)3+3KCl
trước:a-----------b
pư:3b--------------b
sau:a-3b-------------------------b
KOH+Al(OH)3---->KAlO3+H2O
a-3b---------a-3b
mol kết tủa=b+a-3b=a-2b(mol)