cho a gam hỗn hợp A gồm CaO và CaCO3 tác dụng với 300ml dung dịch HCl thu đc 33,3 gam muối CaCl2 và 4480 ml khí CO2 đktc
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1)
a)
$CaO + 2HCl \to CaCl_2 + H_2O$
$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
$n_{CaCO_3} = n_{CO_2} = 0,2(mol)$
$n_{CaCl_2} = 0,3(mol)$
Suy ra:
$n_{CaO} = 0,3 - 0,2 = 0,1(mol)$
$\%m_{CaO} = \dfrac{0,1.56}{0,1.56 + 0,2.100}.100\% = 21,875\%$
$\%m_{CaCO_3} = 78,125\%$
b)
$m_{dd} = 0,1.56 + 0,2.100 + 50 - 0,2.44 = 66,8(gam)$
$C\%_{CaCl_2} = \dfrac{33,3}{66,8}.100\% = 49,85\%$
Câu 4 :
a)
Gọi $n_{Fe} = a(mol) ; n_{MgO} = b(mol)$
Suy ra: $56a + 40b = 19,2(1)$
$Fe + 2HCl \to FeCl_2 + H_2$
$MgO + 2HCl \to MgCl_2 + H_2O$
Theo PTHH : $n_{HCl} = 2a + 2b = 0,4.2 = 0,8(2)$
Từ (1)(2) suy ra a = b = 0,2
$\%m_{Fe} = \dfrac{0,2.56}{19,2}.100\% = 58,33\%$
$\%m_{MgO} = 100\% -58,33\% = 41,67\%$
b)
$n_{FeCl_2} = a = 0,2(mol)$
$n_{MgCl_2} = b = 0,2(mol)$
$m_{muối} = 0,2.127 + 0,2.95 = 44,4(gam)$
Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\n_{CaCl_2}=\dfrac{33,3}{111}=0,3\left(mol\right)\end{matrix}\right.\)
PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2\uparrow+H_2O\)
0,2_____0,4_____0,2____0,2_____0,2 (mol)
\(CaO+2HCl\rightarrow CaCl_2+H_2O\)
0,1_____0,2_____0,1____0,1 (mol)
Ta có: \(\left\{{}\begin{matrix}m_{CaO}=0,1\cdot56=5,6\left(g\right)\\m_{CaCO_3}=0,2\cdot100=20\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{CaO}=\dfrac{5,6}{5,6+20}\cdot100\%=21,875\%\\\%m_{CaCO_3}=78,125\%\end{matrix}\right.\)
Mặt khác: \(m_{CO_2}=0,2\cdot44=8,8\left(g\right)\)
\(\Rightarrow m_{dd\left(sau.p/ứ\right)}=m_{CaO}+m_{CaCO_3}+m_{ddHCl}-m_{CO_2}=66,8\left(g\right)\)
\(\Rightarrow C\%_{CaCl_2}=\dfrac{33,3}{66,8}\cdot100\%\approx49,85\%\)
nCO2=\(\dfrac{4,48}{22,4}=0,2\) mol
nCaCl2=\(\dfrac{33,3}{111}=0,3\)
CaCO2 + 2HCl → CaCl2 + CO2 + H2O
0,2 ← 0,2 ← 0,2
CaO + 2HCl → CaCl2 + H2O
0,1 ← 0,1
a) % CaO=\(\dfrac{0,1.56}{0,1.56+0,2.100}.100\%=21,875\%\)
% CaCO3 =100% - 21,875%= 78,125%
b) a = mCaO+mCaCO3 =0,1.56+0,2.100=25,6g
mdd sau pư= a + mddHCl - mCO2
= 25,6 + 50 - 0,2.44=66,8g
C%CaCl2=\(\dfrac{33,3}{66,8}.100\%\simeq49,85\%\)
PTHH: \(CaO+2HCl\rightarrow CaCl_2+H_2O\) (1)
\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\) (2)
a) Ta có: \(\left\{{}\begin{matrix}\Sigma n_{CaCl_2}=\dfrac{33,3}{111}=0,3\left(mol\right)\\n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}n_{CaO}=0,1mol\\n_{CaCO_3}=0,2mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaO}=0,1\cdot56=5,6\left(g\right)\\m_{CaCO_3}=0,2\cdot100=20\left(g\right)\end{matrix}\right.\)
\(\Rightarrow m_{hh}=20+5,6=25,6\left(g\right)\)
b) Theo các PTHH: \(\left\{{}\begin{matrix}n_{HCl\left(1\right)}=0,2mol\\n_{HCl\left(2\right)}=0,4mol\end{matrix}\right.\)
\(\Rightarrow\Sigma n_{HCl}=0,6mol\) \(\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,3}=2\left(M\right)\)
PT: \(CaO+2HCl\rightarrow CaCl_2+H_2O\) (1)
\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\) (2)
a, Ta có: \(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(\Sigma n_{CaCl_2}=\dfrac{33,3}{111}=0,3\left(mol\right)\)
Theo PT (2): \(n_{CaCl_2}=n_{CO_2}=0,2\left(mol\right)\)
\(\Rightarrow n_{CaCl_2\left(1\right)}=0,3-0,2=0,1\left(mol\right)\)
Theo PT (1): \(n_{CaO}=n_{CaCl_2}=0,1\left(mol\right)\)
Theo PT (2): \(n_{CaCO_3}=n_{CO_2}=0,2\left(mol\right)\)
\(\Rightarrow m_A=m_{CaO}+m_{CaCO_3}=0,1.56+0,2.100=25,6\left(g\right)\)
b, Theo PT (1) + (2): \(\Sigma n_{HCl}=2n_{CaO}+2n_{CaCO_3}=0,6\left(mol\right)\)
\(\Rightarrow C_{M_{ddHCl}}=\dfrac{0,6}{0,3}=2M\)
Bạn tham khảo nhé!
m(hhA) = mNa + mCa + mOH + mCO3 = 8,42(g) (∙)
n(CO2) = nCO3 = 0,03(mol) → mCO3 = 1,8(g)
n(NaCl) = nNa= 0,08(mol) → mNa = 1,84(g)
từ (∙) ta có: mCa + mOH = 4,78(g) <=> 40nCa + 17nOH = 4,78 (1)
áp dụng định luật bảo toàn điện tích cho hhA ta có:
nNa + 2nCa = nOH + 2nCO3 <=> 2nCa - nOH = -0,02 (2)
Từ (1) và (2) → nCa = 0,06(mol)→nCaCl2 = nCa=0,06(mol) → mCaCl2 = 6,66(g) → Đáp án C
Gọi \(\left\{{}\begin{matrix}n_{H_2}=a\left(mol\right)\\n_{CO_2}=b\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}a+b=\dfrac{4,704}{22,4}=0,21\\\overline{M}=\dfrac{2a+44b}{a+b}=12,5.2=25\end{matrix}\right.\)
=> a = 0,095 (mol); b = 0,115 (mol)
Đặt nHCl = x (mol)
\(n_{MgCl_2}=\dfrac{12,825}{95}=0,135\left(mol\right)\)
Bảo toàn Cl: \(n_{CaCl_2}=\dfrac{x-0,27}{2}\left(mol\right)\)
Bảo toàn H: \(n_{H_2O}=\dfrac{x-2.0,095}{2}=\dfrac{x-0,19}{2}\left(mol\right)\)
BTKL:
\(m_{hh\left(bđ\right)}+m_{HCl}=m_{MgCl_2}+m_{CaCl_2}+m_{H_2}+m_{CO_2}+m_{H_2O}\)
=> \(19,02+36,5x=12,825+\dfrac{x-0,27}{2}.111+0,095.2+0,115.44+\dfrac{x-0,19}{2}.18\)
=> x = 0,63 (mol)
=> \(n_{CaCl_2}=0,18\left(mol\right)\)
=> mCaCl2 = 0,18.111 = 19,98 (g)
\(PTHH:CaO+2HCl\rightarrow CaCl2+H2O\)
________0,1____________0,1____________
\(CaCO3+2HCl\rightarrow CaCl2+CO2+H2O\)
0,2_____________0,2_________0,2_________
Ta có :
\(n_{CO2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)
\(n_{CaCl2}=\frac{33,3}{111}=0,3\left(mol\right)\)
\(\rightarrow n_{CaCO3}=0,2\left(mol\right),n_{CaO}=0,1\left(mol\right)\)
\(a=m_{CaCO3}+m_{CaO}=0,2.100+0,1.56=25,6\left(g\right)\)
nCaCl2 = 0.3 mol
nCO2 = 0.2 mol
CaCO3 + 2HCl => CaCl2 + CO2 + H2O
0.2_______0.4______0.2____0.2
CaO + 2HCl => CaCl2 + H2O
0.1_____0.2_____0.1
mA = mCaCO3 + mCaO = 0.2*100 + 0.1*56 = 25.6 g
CM HCl = 0.6/0.3 = 2M