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\(PTHH:Na_2CO_3+2HCl->2NaCl+CO_2+H_2O\)
áp dụng ĐLBTKL ta có
\(m_{Na_2CO_3}+m_{HCl}=m_{NaCl}+m_{CO_2}+m_{H_2O}\\ =>m_{CO_2}=m_{Na_2CO_3}+m_{HCl}-m_{NaCl}-m_{H_2O}\\ =>m_{CO_2}=10,6+7,3-11,7-1,8\\ =>m_{CO_2}=4,4\left(g\right)\)
\(n_{CO_2}=\dfrac{m}{M}=\dfrac{4,4}{44}=0,1\left(mol\right)\\ V_{CO_2\left(dkt\right)}=n\cdot24,79=0,1\cdot24,79=2,479\left(l\right)\)
) Gọi n Na2CO3= x ; n Na2So3 =y
=> 106x + 126y= 55 (I)
Ta có: mHCl =36,5 => nHCl= 1
Na2CO3 + 2HCl --> 2NaCl + CO2 + H2O (1)
x 2x 2x x
Na2SO3 + 2HCl --> 2NaCl + SO2 + H2O (2)
y 2y 2y y
=> 2x + 2y = 1 (II)
(I) và (II) => n Na2CO3= 0,4 và n Na2SO3 = 0,1
(1)=> nCO2 = 0,4
(2) => nSO2= 0,1
=> %V CO2 = 80% => %V SO2 = 20%
b) ΣΣ nNaCl= 1=> mNaCl= 58,5
m dd sau pứ= 55 + 250 - 0,4 x 44 - 0,1 x 64= 281(g)
C% dd NaCl = 20,82%
Lập PTHH theo các sơ đồ sau
CaCO3+2HCl-->CaCl2+H2O+CO2
BaCO3+2HCl-->BaCl2+CO2+H2O
K2CO3+2HBr-->2KBr+CO2+H2O
Na2CO3+2HCl-->2NaCl+CO2+H2O
MgCO3+2HNO3-->Mg(NO3)2+CO2+H2O
NaHCO3+HCl-->NaCl+CO2+H2O
NaHCO3+HNO3-->NaNO3+CO2+H2O
2NaHCO3+H2SO4-->Na2SO4+2CO2+2H2O
2Na2CO3+HNO3-->2NaNO3+CO2+H2O
CuCO3+2HCl-->CuCl2+CO2+H2O
CaCO3+2HCl-->CaCl2+H2O+CO2
BaCO3+2HCl-->BaCl2+CO2+H2O
K2CO3+2HBr-->2KBr+CO2+H2O
Na2CO3+2HCl-->2NaCl+CO2+H2O
MgCO3+2HNO3-->Mg(NO3)2+CO2+H2O
NaHCO3+HCl-->NaCl+CO2+H2O
NaHCO3+HNO3-->NaNO3+CO2+H2O
2NaHCO3+H2SO4-->Na2SO4+2CO2+2H2O
Na2CO3+2HNO3-->2NaNO3+CO2+H2O
CuCO3+2HCl-->CuCl2+CO2+H2O
1) PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
2) Ta có: \(n_{Mg}=\dfrac{1,2}{24}=0,05\left(mol\right)=n_{H_2}\) \(\Rightarrow V_{H_2}=0,05\cdot22,4=1,12\left(l\right)\)
3)
+) Cách 1: Tính theo phương trình
Theo PTHH: \(n_{HCl}=2n_{Mg}=0,1mol\) \(\Rightarrow m_{HCl}=0,1\cdot36,5=3,65\left(g\right)\)
+) Cách 2: Bảo toàn khối lượng
Ta có: \(\left\{{}\begin{matrix}m_{H_2}=0,05\cdot2=0,1\left(g\right)\\m_{MgCl_2}=0,05\cdot95=4,75\left(g\right)\end{matrix}\right.\)
\(\Rightarrow m_{HCl}=m_{MgCl_2}+m_{H_2}-m_{Mg}=4,75+0,1-1,2=3,65\left(g\right)\)
+) Cách 3: Bảo toàn nguyên tố (Bonus)
Theo PTHH: \(n_{MgCl_2}=n_{H_2}=0,05mol\) \(\Rightarrow\left\{{}\begin{matrix}n_{Cl}=0,1mol\\n_H=0,1mol\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Cl}=0,1\cdot35,5=3,55\left(g\right)\\m_H=0,1\cdot1=0,1\left(g\right)\end{matrix}\right.\) \(\Rightarrow m_{HCl}=3,55+0,1=3,65\left(g\right)\)
a, \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH: 2Al + 6HCl → 2AlCl3 + 3H2
Mol: 0,2 0,6 0,2 0,3
\(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)
b, \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)
c, \(m_{ddHCl}=\dfrac{0,6.36,5.100}{10}=219\left(g\right)\)
a) Gọi $n_{CO_2} = a(mol) ; n_{SO_2} = b(mol)$
Ta có :
$a + b = \dfrac{8,96}{22,4} = 0,4(mol)$
$\dfrac{44a + 64b}{a + b} = 27.2$
Suy ra : a = b = 0,2$
$V_{CO_2} = V_{SO_2} = 0,2.22,4 = 4,48(lít)$
b) Theo PTHH : $n_{K_2SO_3} = n_{SO_2} = 0,2(mol)$
$\Rightarrow m_{K_2SO_3} = 0,2.158 = 31,6(gam)$
Gọi $n_{K_2CO_3} = x(mol) ; n_{Na_2CO_3} = y(mol)$
$\Rightarrow 138x + 106y + 31,6 = 56(1)$
$n_{CO_2} = x + y = 0,2(2)$
Từ (1)(2) suy ra : x = y = 0,1
$m_{K_2CO_3} = 0,1.138 = 13,8(gam) ; m_{Na_2CO_3} = 0,1.106 = 10,6(gam)$
PTHH: \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\\ 0,1mol:0,2mol\rightarrow0,2mol:0,1mol:0,1mol\)
\(n_{Na_2CO_3}=\dfrac{10,6}{106}=0,1\left(mol\right)\)
a. \(n_{NaCl}=0,2.58,5=11,7\left(g\right)\)
b. \(V_{CO_2}=0,1.22,4=2,24\left(l\right)\)
c. \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(n_{Na_2CO_3}=\dfrac{m_{Na_2CO_3}}{M_{Na_2CO_3}}=\dfrac{10,6}{106}=0,1\left(mol\right)\)
PTHH:
\(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\)
\(0,1............0,2........0,2.......0,1.......0,1\) (mol)
a. \(m_{NaCl}=M_{NaCl}.n_{NaCl}=0,2.58,5=11,7\left(g\right)\)
b. \(V_{CO_2}=22,4.n_{CO_2}=22,4.0,1=2,24\left(l\right)\)
c. \(m_{HCl}=n_{HCl}.M_{HCl}=0,2.36,5=7,3\left(g\right)\)