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\(nCaCl_2=\dfrac{2,22}{111}=0,02\left(mol\right)\)
\(nAgNO_3=\dfrac{1,7}{170}=0,01\left(mol\right)\)
\(CaCl_2+2AgNO_3\rightarrow2AgCl+Ca\left(NO_3\right)_2\)
1 2 2 1 (mol)
0,005 0,01 0,01 0,005
LTL : \(\dfrac{0,02}{1}>\dfrac{0,01}{2}\)
=> CaCl2 dư , AgNO3 đủ
\(m_{kt}=mAgCl=0,01.143,5=1,435\left(g\right)\)
c1:
\(m_{\left(muối\right)}=m_{Ca\left(NO_3\right)_2}=0,005.164=0,82\left(g\right)\)
c2:
BTKL:
\(mCaCl_{2\left(đủvspứ\right)}=0,005.111=0,555\left(g\right)\)
\(mCaCl_2+mAgNO_3=mAgCl+mCa\left(NO_3\right)_2\)
0,555 + 1,7 = 1,435 + \(mCa\left(NO_3\right)_2\)
\(\Rightarrow mCa\left(NO_3\right)_2=0,555+1,7-1,435=0,82\left(g\right)\)
n H2SO4=\(\dfrac{10\%.490}{2+32+16.4}=0,5mol\)
n Al2O3 =\(\dfrac{10,2}{27.2+16.3}=0,1mol\)
\(Al_2O_3+3H_2SO_4->Al_2\left(SO_4\right)_3+3H_2O\)
bđ 0,1............0,5
pư 0,1............0,3..................0,1
spu 0 ................0,2................0,1
=> sau pư gồm H2SO4 dư , Al2(S04)3 và H2O
m H2SO4 dư = \(0,2.\left(2+32+16.3\right)=19,6g\)
m Al2(SO4)3 = \(0,1\left(27.2+32.3+16.4.3\right)=34,2g\)
m dd = \(490+10,2=500,2g\)
% Al2(SO4)3 = \(\dfrac{34,2}{500,2}.100\sim6,84\%\)
% H2SO4 dư = \(\dfrac{19,6}{500,2}.100\sim3,92\%\)
AgNO3 + HCl \(\rightarrow\)AgCl + HNO3
mHCl=20.\(\dfrac{7,3}{100}=1,46\left(g\right)\)
nHCl=\(\dfrac{1,46}{36,5}=0,04\left(mol\right)\)
Theo PTHH ta có:
nHCl=nAgCl=0,04(mol)
mAgCl=0,04.143,5=5,74(g)
Theo PTHH ta có:
nHCl=nAgNO3=nHNO3=0,04(mol)
mAgNO3=170.0,04=6,8(g)
mdd AgNO3=\(6,8:\dfrac{1,7}{100}=400\left(g\right)\)
mHNO3=63.0,04=2,52(g)
C% dd HNO3=\(\dfrac{2,52}{400+20-5,74}.100\%=0,6\%\)
Ta có nHCl = \(\dfrac{20\times7,3\%}{36,5}\) = 0,04 ( mol )
HCl + AgNO3 \(\rightarrow\) AgCl\(\downarrow\) + HNO3
0,04........0,04.........0,04........0,04
=> mAgCl = 0,04 . 143,5 = 5,74 ( gam )
=> mHNO3 = 63 . 0,04 = 2,52 ( gam )
=> mAgNO3 = 0,04 . 170 = 6,8 ( gam )
=> mdung dịch AgNO3 = 6,8 : 1,7 . 100 = 400 ( gam )
Ta có Mdung dịch = Mtham gia
= 20 + 400 = 420 ( gam )
=> C%HNO3 = \(\dfrac{6,8}{420}\times100\approx\) 1,62 %
nAgNO3 = \(\dfrac{1,7}{170}=0,01\) mol
Pt: CaCl2 + 2AgNO3 --> 2AgCl + Ca(NO3)2
..................0,01 mol----> 0,01 mol
mAgCl = 0,01 . 143,5 = 1,435 (g)
PTHH :
\(2Na+2H_2O-->2NaOH+H_2\)
0,2__________________0,1
\(2NaOH+2AgNO_3-->2NaNO_3+Ag_2O+H_2O\)
0,03______________________________0,015
\(n_{Na}=\dfrac{5,1}{23}=0,2\left(mol\right)\)
\(n_{AgNO_3}=\dfrac{5,1}{170}=0,03\left(mol\right)\)
=>Na dư =>\(n_{NaOH}dư=0,1-0,03=0,07\left(mol\right)\)
=> \(m_{Ag_2O}=0,015.232=3,48\left(g\right)\)