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a)
PTHH :
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
b)
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4}=0,1\left(mol\right)\)
\(\Rightarrow m_{\text{dd}H_2SO_4}=\dfrac{0,1\cdot98}{9,8\%}=100\left(g\right)\)
c)
\(m_{\text{dd}\left(sau\right)}=100+6,5-0,1.2=106,3\left(g\right)\)
\(\Rightarrow C\%_{\text{dd}\left(sau\right)}=\dfrac{0,1\cdot161}{106,3}\cdot100\%=15,15\left(\%\right)\)
a)
NaCl + AgNO3 -> NaNO3 + AgCl↓
0,1 0,1 0,1 0,1
b)
mNaCl = \(\frac{100.5,85}{100}=5,85\)=> nNaCl = 5,85 : 58,5 = 0,1 mol
mAgCl = 0,1 . 143,5 = 14,35 g
c)
mNaNO3 = 0,1 . 85 = 8,5 g
a, NaCl + AgNO3 --> NaNO3 +AgCl
1mol 1mol 1mol 1mol
0,1mol 0,1mol 0,1mol 0,1mol
b, mNaCl=100.5,85%=5,85g
nNaCl=\(\frac{5,85}{58,5}\)=0,1(mol)
mAgNO3=0,1.170=17g
mNaNO3=0,1.85=8,5g
mAgCl=0,1.143,5=14,35g
nH2 = \(\frac{1,68}{22,4}\) = 0,075 (mol)
Mg + 2HCl \(\rightarrow\) MgCl2 + H2\(\uparrow\) (1)
0,075 <--------0,075 <--0,075 (mol)
MgO + 2HCl \(\rightarrow\) MgCl2 + H2O (2)
%mMg= \(\frac{0,075.24}{5,8}\) . 100% = 31,03 %
%m MgO = 68,97%
nMgO = \(\frac{5,8-0,075.24}{40}\) = 0,1 (mol)
Theo pt(2) nMgCl2 = nMgO= 0,1 (mol)
mdd sau pư = 5,8 + 194,35 - 0,075.2 = 200 (g)
C%(MgCl2) = \(\frac{95\left(0,075+0,1\right)}{200}\) . 100% = 8,3125%
Tham khảoa)
BaCl2+H2SO4\(\rightarrow\)BaSO4+2HCl
b)
Quỳ tím có màu đỏ vì có HCl tạo ra nên dd có tính axit
c)
nBaCl2=\(\frac{\text{100.20,8%}}{208}\)=0,1(mol)
nBaSO4=nBaCl2=0,1(mol)
mBaSO4=0,1.233=23,3(g)
d)
nH2SO4=nBaCl2=0,1(mol)
mdd H2SO4=\(\frac{\text{0,1.98}}{4,9\%}\)=200(g)
e)
mdd spu=100+200-23,3=276,7(g)
C%HCl=\(\frac{\text{0,2.36,5}}{276,7}.100\%\)=2,64%
a) H2SO4+BaCl2---->BaSO4+2HCl
b) n\(_{H2SO4}=\frac{200.9,8}{100.98}=0,2\left(mol\right)\)
n\(_{BaCl2}=\frac{800.6,5}{100.208}=0,25\left(mol\right)\)
=> BaCl2 dư
Theo pthh
n\(_{BaSO4}=n_{H2SO4}=0,2\left(mol\right)\)
m\(_{BaSO4}=0,2.233=46,6\left(g\right)\)
m ddsau pư=800+200-46,6=953,4(g)
Theo pthh
n\(_{BaCl2}=n_{H2SO4}=0,2\left(mol\right)\)
n BaCl2 dư=0,25-0,2=0,05(mol)
C% BaCl2=\(\frac{0,05.208}{953,4}.100\%=1,09\%\)
Theo pthh
n\(_{HCl}=2n_{H2SO4}=0,2\left(mol\right)\)
C% HCl=\(\frac{0,2.36,5}{953,4}.100\%=0,77\%\%\)
\(\text{h2so4 + bacl2 = baso4 + h2o}\)
Ta có :
\(\text{n h2so4 = 0,2 mol}\)
\(\text{n bacl2 = 0,25 mol }\)
theo pthh thì n h2so4 = n bacl2
\(\text{mà n bacl2 có > n h2so4}\)
--> h2so4 hết, còn bacl2 dư 0,05 mol
\(\text{m kết tủa = m baso4 = 0,2.233= 46,6g}\)
dd sau pứ là bacl2 dư 0,05mol
\(\text{m dd sau pứ = 200 + 800- 46,6 = 753,4g}\)
\(\text{--> C% Bacl2 = 0,05.208÷753,4.100%= 1,38%}\)
mH2SO4(d2) = 1,7 . 115,29 = 195,993 g
=>mH2SO4 = \(\dfrac{195,993.35}{100}\) = 68,6 g => n = \(\dfrac{68,6}{98}\) = 0,7mol
CuO + H2SO4 -> CuSO4 + H2O
x------>x------->x
Fe3O4 + 4H2SO4 -> Fe2(SO4)3 + FeSO4 +4 H2O
y-------->4y--------->y------------>y
b)ta có : \(\left\{{}\begin{matrix}80x+232y=42,8\\x+4y=0,7\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=0,1mol\\y=0,15mol\end{matrix}\right.\)
%CuO = \(\dfrac{0,1.80}{42,8}\)100% = 18,691%
%Fe3O4 = 100% - 18,691%= 81,309 %
c)md2 = 195,993+42,8 \(\approx\) 238,8 g
C%(CuSO4) = \(\dfrac{0,1.160}{238,8}.100\%\) = 6,7 %
C%(FeSO4) = \(\dfrac{0,15.152}{238,8}.100\%\) = 9,54%
C%(Fe2(SO4)) = \(\dfrac{0,15.400}{238,8}.100\%\) = 25,12%
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)
nH2=4,48/22,4=0,2(mol)
=>nFe=0,2(mol)=>mFe=0,2.56=11,2(g)
=>mFeO=18,4-11,2=7,2(g)
b)nH2SO4=nH2=0,2(mol)
=>mH2SO4 7%=0,2.98=19,6(g)
=>mH2SO4 =19,6:7%=280(g)
c)mFeSO4=0,2.152=30,4(g)
mdd sau pư=18,4+280-0,2.2=298(g)
=>C%FeSO4=\(\frac{30,4}{298}.100\%\)=10,2%
Theo đề bài ta có : ⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩VddH2O4=601,2=50(ml)nNaOH=20.20100.40=0,1(mol){VddH2O4=601,2=50(ml)nNaOH=20.20100.40=0,1(mol)
nFe = 1,68/56 = 0,03 mol
a) Ta có PTHH :
2NaOH + H2SO4 -> Na2SO4 + 2H2O
0,1mol......0,05mol
=> CMH2SO4 = 0,05/0,05=1(M)
sao to vậy