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nNa = 6.9 : 23 = 0.3 mol
4Na + O2 ->2 Na2O
mol : 0.3 -> 0.15
Na2O + H2O -> 2NaOH
mol : 0.15 -> 0.3
mdd = 0.15 x 62 + 140.7 = 150g
C% NaOH = 0.3x40: 150 x 100% = 8%
a, \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
b, Ta có: \(m_{H_2SO_4}=200.9,8\%=19,6\left(g\right)\)
\(\Rightarrow n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)
Theo PT: \(n_{MgO}=n_{MgSO_4}=n_{H_2SO_4}=0,2\left(mol\right)\)
\(\Rightarrow m_{MgO}=0,2.40=8\left(g\right)\)
c, Ta có: m dd sau pư = 8 + 200 = 208 (g)
\(\Rightarrow C\%_{MgSO_4}=\dfrac{0,2.120}{208}.100\%\approx11,54\%\)
\(n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\\ n_{H_2SO_4}=\dfrac{19,6\%.200}{98}=0,4\left(mol\right)\\a, ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\\ b,Vì:\dfrac{0,1}{1}< \dfrac{0,4}{1}\\ \Rightarrow H_2SO_4dư\\ n_{H_2SO_4\left(p.ứ\right)}=n_{ZnSO_4}=n_{ZnO}=0,1\left(mol\right)\\ n_{H_2SO_4\left(dư\right)}=0,4-0,1=0,3\left(mol\right)\\ \Rightarrow m_{H_2SO_4\left(dư\right)}=98.0,3=29,4\left(g\right)\\ c,n_{ZnSO_4}=0,1.161=16,1\left(g\right)\\ m_{ddsau}=m_{ZnO}+m_{ddH_2SO_4}=8,1+200=208,1\left(g\right)\\ \Rightarrow C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{29,4}{208,1}.100\approx14,128\%\\ C\%_{ddZnSO_4}=\dfrac{16,1}{208,1}.100\approx7,737\%\)
ZnO+H2SO4->ZnSO4+H2O
0,1-----0,1-------0,1-------0,1 mol
n ZnO=\(\dfrac{8,1}{81}\)=0,1 mol
m H2SO4 =39,2g =>n H2SO4=\(\dfrac{39,2}{98}\)=0,4 mol
=>H2SO4 , dư 0,3 mol
=>m H2SO4=0,3.98=29,4g
=>C%H2SO4 dư=\(\dfrac{29,4}{200+0,1.18}\).100=14,568%
=>C% ZnSO4=\(\dfrac{0,1.161}{200+0,1.18}.100=7,9781\%\)
Fe+2HCl->FeCl2+H2
0,125--0,25---0,125-0,125
m HCl=9,125 g=>n HCl=\(\dfrac{9,125}{26,5}\)=0,25 mol
=>m Fe=0,125.56=7g
=>VH2=0,125.22,4=2,8l
=>C%FeCl2=\(\dfrac{0,125.127}{7+182,5-0,25}\).100=8,388%
a.b.\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1mol\)
\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)
0,1 0,3 0,2 ( mol )
\(V_{H_2}=0,3.22,4=6,72l\)
c.\(n_{HCl}=\dfrac{125.14,6\%}{36,5}=0,5mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,2 < 0,5 ( mol )
0,2 0,2 0,2 ( mol )
\(m_{FeCl_2}=0,2.127=25,4g\)
\(m_{ddspứ}=\left(0,2.56\right)+125-0,2.2=135,8g\)
\(C\%_{FeCl_2}=\dfrac{25,4}{135,8}.100=18,7\%\)
\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\:\right)\\
Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
0,1 0,3 0,2
=> \(m_{H_2}=0,3.22,4=6,72\left(l\right)\)
\(m_{HCl}=125.14,6\%=18,25\left(g\right)\)
\(m_{Fe}=0,2.56=11,2\left(g\right)\)
Fe + 2HCl →FeCl2 +H2
\(C\%=\dfrac{11,2}{18,25}.100\%=61,3\%\)
\(a) Fe + 2HCl \to FeCl_2\\ b) n_{HCl} = \dfrac{182,5.5\%}{36,5} = 0,25(mol)\\ n_{FeCl_2} = n_{H_2} = n_{Fe} = \dfrac{1}{2}n_{HCl} = 0,125(mol)\\ \Rightarrow m_{Fe} = 0,125.56 = 7(gam) ; V = 0,125.22,4 = 2,8(lít)\\ c) m_{dd\ sau\ phản\ ứng} = m_{Fe} + m_{dd\ HCl} - m_{H_2} = 7 + 182,5 - 0,125.2 = 189,25(gam)\\ C\%_{FeCl_2} = \dfrac{0,125.127}{189,25}.100\% = 8,39\%\)
a)b)c)d) mBaCl2=150.16,64%=24,96g
=>nBaCl2=0,12 mol
mH2SO4=100.14,7%=14,7g=>nH2SO4=0,15mol
BaCl2 + H2SO4 =>BaSO4 +2HCl
Bđ: 0,12 mol; 0,15 mol
Pứ: 0,12 mol=>0,12 mol=>0,12 mol=>0,24 mol
Dư: 0,03 mol
Dd ban đầu chứa BaCl2 0,12 mol và H2SO4 0,15 mol
Dd A sau phản ứng chứa HCl 0,24 mol và H2SO4 dư 0,03 mol
mHCl=0,24.36,5=8,76g
mH2SO4=0,03.98=2,94g
Kết tủa B là BaSO4 0,12 mol=>mBaSO4=0,12.233=27,96g
mddA=mddBaCl2+mddH2SO4-mBaSO4
=150+100-27,96=222,04g
C%dd HCl=8,76/222,04.100%=3,945%
C% dd H2SO4=2,94/222,04.100%=1,324%
e) HCl +NaOH =>NaCl +H2O
0,24 mol=>0,24 mol
H2SO4 +2NaOH =>Na2SO4 + 2H2O
0,03 mol=>0,06 mol
TÔNG nNaOH=0,3 mol
=>V dd NaOH=0,3/2=0,15 lit
\(n_{P_2O_5}=\dfrac{99,4}{142}=0,7\left(mol\right)\)
\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)
0,7 2,1 1,4
a, \(m_{H_3PO_4}=1,4.98=137,2\left(g\right)\)
\(m_{ddH_3PO_4}=99,4+500=599,4\left(g\right)\)
Kl nước trong dd A :
\(m_{H_2O}=599,4-137,2=462,2\left(g\right)\)
\(b,C\%_{H_3PO_4}=\dfrac{137,2}{599,4}.100\%\approx22,89\%\)
\(c,C_M=\dfrac{n}{V}=\dfrac{1,4}{0,5}=2,8M\)
Na2O+H2O->2NaOH (hiện tg hóa học )
0,05---------------0,1 mol
n Na2O=0,05 mol
=>dd chưa NaOH
=>C%=\(\dfrac{0,1.40}{3,1+50}\).100=7,53%
có xảy ra hiện tượng hóa học : Na2O tan trong nước
ct : Na2O
mdd = 3,1+50 = 53,1(g)
\(C\%=\dfrac{3,1}{53,1}.100\%=5,838\%\)