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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,mdd 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)\(m_{CuSO_4}=0,3125.160=50\left(g\right)\)
\(m_{ddCuSO_4}=20+156,25=176,25\left(g\right)\)
\(C\%_{CuSO_4}=\dfrac{50}{176,25}.100\approx28,37\%\)
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%
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, \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
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: \(n_{H_2SO_4}=n_{Mg}+n_{MgO}=0,15\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,15.98=14,7\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{14,7}{19,6\%}=75\left(g\right)\)
Tóm tắt
\(V_{H_2\left(đktc\right)}=8,96l\\ C_{\%H_2SO_4}=19,6\%\\ a)m_{Zn}=?\\ m_{ddH_2SO_4}=?\\ b)C_{\%ZnSO_4}=?\)
\(a)n_{H_2}=\dfrac{8,96}{22,4}=0,4mol\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
0,4 0,4 0,4 0,4
\(m_{Zn}=0,4.65=26g\\ m_{ddH_2SO_4}=\dfrac{0,4.98}{19,6}\cdot100=200g\\ b)C_{\%ZnSO_4}=\dfrac{0,4.161}{26+200-0,4.2}\cdot100=28,6\%\)
a: \(n_{H_2SO_4}=0.25\cdot2=0.5\left(mol\right)\)
\(n_{Al_2O_3}=\dfrac{10.2}{27\cdot2+16\cdot3}=0.1\left(mol\right)\)
\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
0,1 0,5
Vì 0,1/1<0,5/3
nên Al2O3 hết, H2SO4 dư
=>Tính theo Al2O3
b:
\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
0,1 0,3 0,1
\(m_{H_2SO_4\left(pư\right)}=0.3\cdot98=29.4\left(g\right)\)
\(m_{muối}=0.1\left(54+3\cdot96\right)=34.2\left(g\right)\)
\(n_{Fe2O3}=\dfrac{4,8}{160}=0,03\left(mol\right)\)
Pt ; \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O|\)
1 3 1 3
0,03 0,09 0,03
a) \(n_{H2SO4}=\dfrac{0,03.3}{1}=0,09\left(mol\right)\)
\(m_{H2SO4}=0,09.98=8,82\left(g\right)\)
\(m_{ddH2SO4}=\dfrac{8.82.100}{19,6}=45\left(g\right)\)
b) \(n_{Fe2\left(SO4\right)3}=\dfrac{0,09.1}{3}=0,03\left(mol\right)\)
⇒ \(m_{Fe2\left(SO4\right)3}=0,03.400=12\left(g\right)\)
\(m_{ddspu}=4,8+45=49,8\left(g\right)\)
\(C_{Fe2\left(SO4\right)3}=\dfrac{12.100}{49,8}=24,1\)0/0
Chúc bạn học tốt