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a) Gọi \(\left\{{}\begin{matrix}n_{MgCO_3}=a\left(mol\right)\\n_{CaCO_3}=b\left(mol\right)\end{matrix}\right.\)
\(n_{CO_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)
PTHH:
MgCO3 + 2CH3COOH ---> (CH3COO)2Mg + CO2 + H2O
a---------->2a-------------------->a----------------->a
CaCO3 + 2CH3COOH ---> (CH3COO)2Ca + CO2 + H2O
b---------->2b------------------>b------------------>b
=> \(\left\{{}\begin{matrix}84a+100b=31,8\\a+b=0,35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,15\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}m_{MgCO_3}=0,2.84=16,8\left(g\right)\\m_{CaCO_3}=0,15.100=15\left(g\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\%m_{MgCO_3}=\dfrac{16,8}{31,8}.100\%=52,8\%\\\%m_{CaCO_3}=100\%-52,8\%=47,2\%\end{matrix}\right.\)
b) \(V_{dd}=\dfrac{2.0,2+2.0,15}{2}=0,35\left(l\right)\)
c) \(m_{muối}=0,2.142+0,15.158=52,1\left(g\right)\)
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: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
a) Ta có: \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)=n_{Mg}\)
\(\Rightarrow\%m_{Mg}=\dfrac{0,5\cdot24}{16}=75\%\) \(\Rightarrow\%m_{MgO}=25\%\)
b) Ta có: \(\left\{{}\begin{matrix}n_{Mg}=0,5\left(mol\right)\\n_{MgO}=\dfrac{16\cdot25\%}{40}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{HCl}=2n_{Mg}+2n_{MgO}=1,2\left(mol\right)\) \(\Rightarrow m_{ddHCl}=\dfrac{1,2\cdot36,5}{20\%}=219\left(g\right)\)
c) Theo các PTHH: \(\left\{{}\begin{matrix}n_{H_2}=0,5\left(mol\right)\\n_{MgCl_2}=0,6\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{H_2}=0,5\cdot2=1\left(g\right)\\m_{MgCl_2}=0,6\cdot95=57\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{hhA}+m_{ddHCl}-m_{H_2}=234\left(g\right)\) \(\Rightarrow C\%_{MgCl_2}=\dfrac{57}{234}\cdot100\%\approx24,36\%\)
Cho mình hỏi ở cái PTHH ấy! sao ta không tính số mol ở dưới??
\(Đặt:\)
\(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\)
\(m_{hh}=24x+56y=13.6\left(g\right)\\ n_{H_2}=x+y=0.3\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x=0.1\\y=0.2\end{matrix}\right.\)
\(\%Mg=\dfrac{0.1\cdot24}{13.6}\cdot100\%=17.64\%\\ \%Fe=100-17.64=82.36\%\)
\(n_{HCl}=2n_{H_2}=2\cdot0.3=0.6\left(mol\right)\)
\(V_{HCl}=\dfrac{0.6}{2}=0.3\left(l\right)\)
\(m_Y=m_{MgCl_2}+m_{FeCl_2}=0.1\cdot95+0.2\cdot127=34.9\left(g\right)\)
a) MgO + 2HCl ---> MgCl2 + H2O (1)
MgCO3 + 2HCl ---> MgCl2 + H2O + CO2 (2)
b) \(n_{CO_2}=\dfrac{m}{M}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
=> \(m_{MgCO_3}=n.M=0,1.84=8,4\left(g\right)\)
\(m_{MgO}=16-8,4=8\left(g\right)\)
c) \(n_{MgO}=\dfrac{m}{M}=\dfrac{8}{40}=0,2\left(mol\right)\)
=> \(n_{HCl\left(1\right)}=0,4\left(mol\right)\); nHCl(2) = 0,2(mol)
=> nHCl = 0.4 + 0,2 = 0,6 (mol)
=> VHCl = \(\dfrac{n}{C_M}=\dfrac{0,6}{1,5}=0,4\left(l\right)=400\left(ml\right)\)
c)
MgCO3 + 2HCl-----> MgCl2 + H2O +CO2
CaCO3 + 2HCl----> CaCl2 + h2O +CO2
Áp dụng định luật bảo toàn khối lượng
m muối khan=mX + mHCl-m H2O -mCO2= 63,6 + 1,4*36,5-0,7*18-0,7*44=71,3g
bạn kiểm tra lại thử nha
chỉ có vậy thô ìa , mình tưởng phức tapj lắm chứ , cảm ơn bạn nhé