a,\(n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\)

PTHH: CuO + H₂SO₄ →CuSO₄ + H₂O

Mol:     0,25      0,25         0,25

\(m_{ddH_2SO_4}=\dfrac{0,25.98.100}{19,6}=125\left(g\right)\)

b,m_{dd sau pứ} = 20+125 = 145 (g)

\(C\%_{ddCuSO_4}=\dfrac{0,25.160.100\%}{145}=27,59\%\)

\(Cu+H_2SO_4\rightarrow CuSO_4+H_2\)
0,3125    0,3125   0,3125        (mol)
a)\(n_{Cu}=\dfrac{20}{64}=0,3125\left(mol\right)\)
\(m_{H_2SO_4}=0,3125.98=30,625\left(g\right)\)
\(m_{ddH_2SO_4}=\dfrac{30,625}{19,6}.100=156,25\left(g\right)\)
b)\(m_{CuSO_4}=0,3125.160=50\left(g\right)\)
\(m_{ddCuSO_4}=20+156,25=176,25\left(g\right)\)
\(C\%_{CuSO_4}=\dfrac{50}{176,25}.100\approx28,37\%\)

\(n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ a.0,25.........0,25..........0,25.......0,25\left(mol\right)\\ Đặt:a=m_{ddH_2SO_4}\left(g\right)\Rightarrow m_{ddX}=a+20\left(g\right)\\Vì.axit.dư\Rightarrow m_{CuSO_4}=0,25.160=40\left(g\right)\\ \dfrac{40}{a+20}.100\%=16\%\Leftrightarrow a=230\left(g\right)\\ \Rightarrow m_{ddH_2SO_4}=230\left(g\right)\\ b.m_{H_2SO_4\left(dư\right)}=230.19,6\%-0,25.98=20,58\left(g\right)\\ C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{20,58}{230+20}.100=8,232\%\)

mH2SO4= 36(g) -> nH2SO4=18/49(mol)

a) PTHH: Fe2O3 + 3 H2SO4 -> Fe2(SO4)3 + 3 H2O

nFe2(SO4)3= nFe2O3= nH2SO4/3= 18/49 : 3=6/49(mol)

=>mFe2O3=6/49 . 160=960/49 (g)

b) mFe2(SO4)3= 400. 6/49=2400/49(g)

mdd(sau)= mFe2O3+ mddH2SO4= 960/49 + 50= 3410/49

=> C%ddFe2(SO4)3= [ (2400/49)/ (3410/49)].100=70,381%

=> C%ddFe2(SO4)3= (48,98/

mKOH=28(g)

nKOH=0.5(mol)

PTHH:2KOH+H2SO4->K2SO4+2H2O

a)Theo pthh:nH2SO4=1/2 nKOH->nH2SO4=0.25(mol)

mH2SO4=0.25*98=24.5(g)

C%ddH2SO4=24.5/100*100=24.5%

theo pthh:nK2SO4=nH2SO4->nK2SO4=0.25(mol)

mK2SO4=0.25*(39*2+96)=43.5(g)

c)mdd sau phản ứng:200+100=300(g)

d) C% muối=43.5:300*100=14.5%

dd la j bn

a, \(n_{Fe}=\dfrac{0,56}{56}=0,01\left(mol\right)\)

PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

Theo PT: \(n_{FeSO_4}=n_{H_2}=n_{Fe}=0,01\left(mol\right)\)

\(\Rightarrow m_{FeSO_4}=0,01.152=1,52\left(g\right)\)

\(V_{H_2}=0,01.22,4=0,224\left(l\right)\)

b, \(n_{H_2SO_4}=n_{Fe}=0,01\left(mol\right)\Rightarrow m_{ddH_2SO_4}=\dfrac{0,01.98}{19,6\%}=5\left(g\right)\)

c, Ta có: m _{dd sau pư} = 0,56 + 5 - 0,01.2 = 5,54 (g)

\(\Rightarrow C\%_{FeSO_4}=\dfrac{1,52}{5,54}.100\%\approx27,44\%\)

Fe2O3 +3H2SO4----.Fe2(SO4)3 +3H2O

a) Ta có

n\(_{Fe2O3}=\frac{4}{160}=0,025\left(mol\right)\)

Theo pthh

n\(_{H2SO4}=3n_{Fe}=0,075\left(mol\right)\)

m\(_{H2SO4}=0,075.98=7,35\left(g\right)\)

b)m\(_{ddH2SO4}=\frac{7,35.100}{9,8}=75\left(g\right)\)

c) Theo pthh

n\(_{Fe2\left(SO4\right)3}=n_{Fe}=0,025\left(mol\right)\)

m\(_{Fe2\left(SO4\right)3}=0,025.400=10\left(g\right)\)

C%=\(\frac{10}{75+4}=12,66\%\)

Chúc bạn học tốt

a)

$n_{Zn} = \dfrac{13}{65} = 0,2(mol) ; n_{H_2 SO_4} = 0,5.2 = 1(mol)$
$Zn + H_2SO_4 \to ZnSO_4 + H_2$
Ta thấy : 

$n_{Zn} < n_{H_2SO_4}$ nên $H_2SO_4$ dư

$n_{ZnSO_4} = n_{H_2SO_4\ pư} = n_{Zn} = 0,2(mol)$
$m_{ZnSO_4} = 0,2.161=32,2(gam)$
$m_{H_2SO_4\ pư} = 0,2.98 = 19,6(gam)$

b)

$n_{H_2SO_4\ dư} = 1 - 0,2 = 0,8(mol)$
$C_{M_{H_2SO_4\ dư}} = \dfrac{0,8}{0,5} = 1,6M$
$C_{M_{FeSO_4}} = \dfrac{0,2}{0,5} = 0,4M$

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{H_2SO_4}=0,5.2=1\left(mol\right)\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ a.Vì:\dfrac{0,2}{1}< \dfrac{1}{1}\Rightarrow H_2SO_4dư\\ n_{H_2SO_4\left(p.ứ\right)}=n_{ZnSO_4}=n_{Zn}=0,2\left(mol\right)\\ m_{H_2SO_4\left(p.ứ\right)}=0,2.98=19,6\left(g\right)\\ m_{ZnSO_4}=161.0,2=32,2\left(g\right)\\ b.V_{ddsau}=V_{ddH_2SO_4}=0,5\left(l\right)\\ C_{MddZnSO_4}=\dfrac{0,2}{0,5}=0,4\left(M\right)\\ C_{MddH_2SO_4\left(dư\right)}=\dfrac{1-0,2}{0,5}=1,6\left(M\right)\)

a, \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

b, \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Theo PT: \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Mg}=0,1.24=2,4\left(g\right)\)

\(\Rightarrow m_{MgO}=4,4-2,4=2\left(g\right)\)

c, \(n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{Mg}+n_{MgO}=0,15\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,15.98=14,7\left(g\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{14,7}{19,6\%}=75\left(g\right)\)