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\(n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\\ n_{H_2SO_4}=\dfrac{19,6\%.200}{98}=0,4\left(mol\right)\\a, ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\\ b,Vì:\dfrac{0,1}{1}< \dfrac{0,4}{1}\\ \Rightarrow H_2SO_4dư\\ n_{H_2SO_4\left(p.ứ\right)}=n_{ZnSO_4}=n_{ZnO}=0,1\left(mol\right)\\ n_{H_2SO_4\left(dư\right)}=0,4-0,1=0,3\left(mol\right)\\ \Rightarrow m_{H_2SO_4\left(dư\right)}=98.0,3=29,4\left(g\right)\\ c,n_{ZnSO_4}=0,1.161=16,1\left(g\right)\\ m_{ddsau}=m_{ZnO}+m_{ddH_2SO_4}=8,1+200=208,1\left(g\right)\\ \Rightarrow C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{29,4}{208,1}.100\approx14,128\%\\ C\%_{ddZnSO_4}=\dfrac{16,1}{208,1}.100\approx7,737\%\)
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=\(\dfrac{0,1.161}{200+0,1.18}.100=7,9781\%\)
1
\(a)m_{H_2O}=250-5=245g\\b )C_{\%NaCl}=\dfrac{5}{250}\cdot100=2\%\)
\(2\\ m_{ddCuSO_4}=\dfrac{15.100}{5}=300g\\ m_{H_2O}=300-15=285g\)
Câu 1:
a, Ta có: m dd = m chất tan + mH2O ⇒ mH2O = 250 - 5 = 245 (g)
b, \(C\%_{NaCl}=\dfrac{5}{250}.100\%=2\%\)
Câu 2:
Ta có: \(C\%_{CuSO_4}=\dfrac{15}{m_{ddCuSO_4}}.100\%=5\%\)
\(\Rightarrow m_{ddCuSO_4}=300\left(g\right)\)
⇒ mH2O = 300 - 15 = 285 (g)
a, nFe = 0,56/56 = 0,01 (mol)
PTHH: Fe + H2SO4 -> FeSO4 + H2
Mol: 0,01 ---> 0,01 ---> 0,01 ---> 0,01
mFeSO4 = 0,01 . 152 = 1,52 (g)
VH2 = 22,4 . 0,01 = 0,224 (l)
b, mH2SO4 = 0,01 . 98 = 0,98 (g)
c, mddH2SO4 = 0,98/19,6% = 5 (g)
d, mdd (sau p/ư) = 5 + 0,56 = 5,56 (g)
C%FeSO4 = 1,52/5,56 = 27,33%
a, nFe = 0,56/56 = 0,01 (mol)
PTHH: Fe + H2SO4 -> FeSO4 + H2
Mol: 0,01 ---> 0,01 ---> 0,01 ---> 0,01
mFeSO4 = 0,01 . 152 = 1,52 (g)
VH2 = 22,4 . 0,01 = 0,224 (l)
b, mH2SO4 = 0,01 . 98 = 0,98 (g)
c, mddH2SO4 = 0,98/19,6% = 5 (g)
d, mdd (sau p/ư) = 5 + 0,56 = 5,56 (g)
C%FeSO4 = 1,52/5,56 = 27,33%
\(a,m_{NaCl}=\dfrac{8.200}{100}=16\left(g\right)\\ b,m_{HCl}=\dfrac{250.14}{100}=35\left(g\right)\\ m_{H_2SO_4}=\dfrac{19,6.300}{100}=58,8\left(g\right)\)
a)
\(m_{H_2SO_4}=\dfrac{300.19,6}{100}=58,8\left(g\right)\)
=> \(m_{dd.H_2SO_4.9,8\%}=\dfrac{58,8.100}{9,8}=600\left(g\right)\)
=> \(m_{H_2O\left(thêm\right)}=600-300=300\left(g\right)\)
b)
\(n_{HCl}=0,2.2=0,4\left(mol\right)\)
=> \(V_{dd.HCl.1,5M}=\dfrac{0,4}{1,5}=\dfrac{4}{15}\left(l\right)\)
=> \(V_{H_2O\left(thêm\right)}=\dfrac{4}{15}-0,2=\dfrac{1}{15}\left(l\right)=\dfrac{200}{3}\left(ml\right)\)
=> \(m_{H_2O\left(thêm\right)}=\dfrac{200}{3}.1=\dfrac{200}{3}\left(g\right)\)
\(n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{19,6}{2+32+16\cdot4}=0,2\left(mol\right)\\ PTHH:Zn+H_2SO_4->ZnSO_4+H_2\)
tỉ lệ: 1 : 1 : 1 : 1
n(mol) 0,2<---0,2------>0,2-------->0,2
\(m_{ZnSO_4}=n\cdot M=0,2\cdot\left(65+32+16\cdot4\right)=32,2\left(g\right)\\ V_{H_2\left(dktc\right)}=n\cdot22,4=0,2\cdot22,4=4,48\left(l\right)\)
a) \(m_{NaCl}=\dfrac{8.200}{100}=16\left(g\right)\)
b) \(m_{HCl}=\dfrac{14.250}{100}=35\left(g\right)\)
c) \(m_{H_2SO_4}=\dfrac{19,6.300}{100}=58,8\left(g\right)\)
\(n_{Fe_2O_3}=\dfrac{24}{160}=0,15(mol)\\ n_{H_2SO_4}=\dfrac{250.19,6\%}{100\%.98}=0,5(mol)\\ PTHH:Fe_2O_3+3H_2SO_4\to Fe_2(SO_4)_3+3H_2O\)
Vì \(\dfrac{n_{Fe_2O_3}}{1}<\dfrac{n_{H_2SO_4}}{3}\) nên \(H_2SO_4\) dư
\(\Rightarrow n_{Fe_2(SO_4)_3}=n_{Fe_2O_3}=0,15(mol)\\ \Rightarrow m_{Fe_2(SO_4)_3}=0,15.400=60(g)\)
nFe2O3=\(\dfrac{24}{160}\)=0,15(mol)
nH2SO4=\(\dfrac{250.19,6\%}{100\%.98}\)=0,5(mol)
PTHH:Fe2O3+3H2SO4→Fe2(SO4)3+3H2O
=>H2SO4 dư
⇒nFe2(SO4)3=nFe2O3=0,15(mol)
⇒mFe2(SO4)3=0,15.400=60(g)
\(n_{Fe_2O_3}=\dfrac{24}{160}=0,15\left(mol\right)\\ m_{H_2SO_4}=19,6\%.250=49\left(g\right)\\ \rightarrow n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)
PTHH: Fe2O3 + 3H2SO4 ---> Fe2(SO4)3 + 3H2O
LTL: \(\dfrac{0,15}{1}< \dfrac{0,5}{3}\rightarrow\) H2SO4 dư
Theo pthh: \(n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,15\left(mol\right)\)
\(\rightarrow m_{Fe_2\left(SO_4\right)_3}=0,15.400=60\left(g\right)\)
Ta có: \(n_{ZnO}=\dfrac{16,2}{81}=0,2\left(mol\right)\)
\(m_{H_2SO_4}=250.19,6\%=49\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)
PT: \(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)
Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,5}{1}\), ta được H2SO4 dư.
Theo PT: \(n_{ZnSO_4}=n_{ZnO}=0,2\left(mol\right)\)
\(\Rightarrow m_{ZnSO_4}=0,2.161=32,2\left(g\right)\)
\(PTPU:ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)
\(0,2:0,2:0,2:0,2\left(mol\right)\)
\(n_{ZnO}=\dfrac{m}{M}=\dfrac{16,2}{81}=0,2\left(mol\right)\)
\(m_{ZnSO_4}=n.M=0,2.161=32,2\left(g\right)\)