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a) 2AgNO3+CaCl2---->2AgCl+Ca(NO3)2
n AgNO3=1,7/170=0,01(mol)
n CaCl2=2,22/111=0,02(mol)
----> CaCl2 dư
Theo pthh
n AgCl=n AgNO3=0,01(mol)
m AgCl=0,01.143,5=14,35(g)
V dd sau pư=70+30=`100ml=0,1(l)
n CaCl2 dư=0,02-0,005=0,015(mol)
CM CaCl2=0,015/0,1=0,15(M)
Theo pthh
n Ca(NO3)2=1/2 n AgCl=0,005(mol)
CM Ca(NO3)2=0,005/0,1=0,05(M)
Bài 2
BaCl2+H2SO4--->BaSO4+2HCl
a) n BaCl2=400.5,2/100=20,8(g)
n BaCl2=20,8/208=0,1(mol)
m H2SO4=100.1,14.20/100=22,8(g)
n H2SO4=22,8/98=0,232(mol)
---->H2SO4 dư
Theo pthh
n BaSO4=n BaCl2=0,1(mol)
m BaSO4=0,1.233=23,3(g)
b) m dd sau pư=400+114-23,3
=490,7(g)
Theo pthh
n HCl=2n BaCl2=0,2(mol)
C%HCl=\(\frac{0,2.36,5}{490,7}.100\%=1,88\%\)
n H2SO4 dư=0,232-0,1=0,132(mol)
C% H2SO4=\(\frac{0,132.98}{490,7}.100\%=2,64\%\)
B1:
\(n_{AgNO3}=0,01\left(mol\right);n_{CaCl2}=0,2\left(mol\right)\)
PTHH:\(2AgNO3+CaCl2\rightarrow2AgCl2\downarrow+Ca\left(NO3\right)2\)
Trước :0,01................0,02..........................................................(mol)
Pứng:\(0,01\rightarrow0,005\rightarrow0,01\rightarrow0,005\)
Dư: 0............................0,015......................................................(mol)
\(m\downarrow_{AgCL}=0,01.143,5=1,435\left(g\right)\)
Trong dd sau phản ứng chứa: \(\left\{{}\begin{matrix}Ca\left(NO3\right)2:0,005\left(mol\right)\\CaCl2:0,015\left(mol\right)\end{matrix}\right.\)
\(C_{M_{Ca\left(NO3\right)2}}=\frac{0,005}{0,1}=0,05M\)
\(C_{M_{CaCl2}}=\frac{0,015}{0,1}=0,15M\)
Bài 2:\(n_{BaCl2}=\frac{400.5,2}{100.208}=0,1\left(mol\right)\)
\(D=\frac{m_{dd}}{v_{dd}};C\%=\frac{m_{ct}}{m_{dd}}.100\Rightarrow m_{H2SO4}=\frac{D.v.d^2.C\%}{100}=22,8g\)
\(\Rightarrow n_{H2SO4}=0,23\left(mol\right)\)
\(BaCl2+HSO4\rightarrow BaSO4\downarrow+2HCl\)
0,1..............0,1............0,1.................0,2.....(mol)
\(a,m_{\downarrow}=0,1.223=23,3\left(g\right)\)
\(b,m_{dd_{saupu}}=m_{BaCl2}+m_{dd_{H2SO4}}-m_{\downarrow}_{BaSO4}\)
\(=400+1,14.100-23,3=490,7\)
\(\Rightarrow C\%_{HCl}=\frac{0,2.36,5}{490,7}.100\%=1,48\%\)
\(\%H2SO4_{du}=\frac{\left(0,23-0,1\right).98}{490,7}.100=2,59\%\)
1.
Sửa đề : 8g kết tủa
Đặt:
nNa2CO3= x mol
nNaHCO3= y mol
mhh= 106x + 84y = 7.6g (1)
Na2CO3 + 2HCl --> 2NaCl + CO2 + H2O
x__________2x____________x
NaHCO3+ HCl --> NaCl + CO2 + H2O
y_________y____________y
nCaCO3= 8/100= 0.08 mol
Ca(OH)2 + CO2 -->CaCO3 + H2O
__________0.08_____0.08
nCO2= x + y = 0.08 (2)
Giải (1) và (2) :
x=y= 0.04
mNa2CO3= 0.04*106= 4.24g
mNaHCO3= 3.36g
%Na2CO3= 55.79%
%NaHCO3= 44.21%
b)
nHCl = 2x + y = 0.12 mol
mHCl = 0.12*36.5=4.38g
mddHCl = 4.38*100/20=21.9g
VddHCl = mdd/D= 21.9/1.14= 19.21ml
2, Gọi kim loại cần tìm là A có hóa trị là n
CTTQ : A2(CO3)n
A\(_2\)(CO3)\(_n\) + nH\(_2\)SO\(_4\) → A\(_2\)(SO\(_4\))\(_n\) + nCO\(_2\) + nH\(_2\)O
(mol) 1.................n........................1..................n
m(H2SO4) = 98n (gam) => m(dd H2SO4) = \(\frac{98n\cdot100\%}{10\%}\) = 980n (gam)
m(CO2) = 44n (gam)
m(dd sau ) = m( muối bđ) + m(dd H2SO4) - m(CO2)
= 1*(2A + 60n) + 980n - 44n
= 2A + 996n
Ta lại có :
\(\frac{1\cdot\left(2A+96n\right)}{2A+996n}\cdot100\%\) = 13,63 %
<=> 200A + 9600n = 27,26A + 13575,48n
<=> 172,24A = 3975,48n
<=> A = 23n
Với n = 1 thì A = 23 (Na) (thỏa mãn )
Vậy công thức của muối ban đầu là Na2CO3
Xét \(A=\frac{n_{NaOH}}{n_{CO2}}=\frac{b}{a}\)
=> 1<A<2 => ta có:
PTHH: NaOH + CO2 --> NaHCO3
_______ a <------ a --------> a ______(mol)
=> \(n_{NaOH\left(dư\right)}=b-a\left(mol\right)\)
NaOH dư sẽ pư với NaHCO3
NaOH + NaHCO3 --> Na2CO3 + H2O
(b-a) --> (b-a) ----------> (b-a)________(mol)
=> \(\left\{{}\begin{matrix}n_{NaHCO3}=a-\left(b-a\right)=2a-b\left(mol\right)\\n_{Na2CO3}=b-a\left(mol\right)\end{matrix}\right.\)
- P1:
PTHH: 2NaHCO3 + CaCl2 --> CaCO3 + 2NaCl + Co2 + H2O
______ (2a-b) --------------------> (a-0,5b)__________________ (mol)
Na2CO3 + CaCl2 --> 2NaCl + CaCO3
(b-a) -------------------------------> (b-a) (mol)
=> \(n_{CaCO3}=\left(a-0,5b\right)+\left(b-a\right)=0,5.b\left(mol\right)\)
=> \(m_1=m_{CaCO3}=100.0,5b=50b\left(g\right)\)
- P2:
PTHH: 2NaHCO3 + Ba(OH)2 ---> Na2CO3 + BaCO3 + 2H2O
_______ (2a-b) -----------------------------------> (a-0,5b)______(mol)
______ Na2CO3 + Ba(OH)2 --> BaCO3 + 2NaOH
________(b-a) ---------------------> (b-a)__________(mol)
=> \(n_{BaCO3}=\left(a-0,5b\right)+\left(b-a\right)=0,5b\left(mol\right)\)
=> \(m_{BaCO3}=197.0,5b=98,5.b\left(g\right)\)
Chắc z 