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a) $CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
Theo PTHH :
$n_{CO_2} = n_{CaCO_3} = \dfrac{10}{100} = 0,1(mol)$
$V_{CO_2} = 0,1.22,4 = 2,24(lít)$
b) $n_{HCl} = 2n_{CaCO_3} = 0,2(mol)$
$C_{M_{HCl}} = \dfrac{0,2}{0,25} = 0,8M$
c) $CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O$
$n_{CaCO_3} = n_{CO_2} = 0,1(mol)$
$m_{CaCO_3} = 0,1.100 = 10(gam)$
a, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Theo PT: \(n_{H_2}=n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b, \(m_{FeCl_2}=0,2.127=25,4\left(g\right)\)
c, \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,3}=\dfrac{4}{3}\left(M\right)\)
\(C_{M_{FeCl_2}}=\dfrac{0,2}{0,3}=\dfrac{2}{3}\left(M\right)\)
a. PTHH: 2NaOH + CO2 ---> Na2CO3 + H2O
b. Ta có: \(C_{M_{NaOH}}=\dfrac{n_{NaOH}}{500:1000}=0,5M\)
=> nNaOH = 0,25(mol)
Theo PT: \(n_{Na_2CO_3}=\dfrac{1}{2}.n_{NaOH}=\dfrac{1}{2}.0,25=0,125\left(mol\right)\)
Do thể tích dung dịch không đổi sau phản ứng nên:
\(V_{dd_{Na_2CO_3}}=V_{NaOH}=500:1000=0,5\left(lít\right)\)
=> \(C_{M_{Na_2CO_3}}=\dfrac{0,125}{0,5}=0,25M\)
PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)
Ta có; \(n_{CaCO_3}=\dfrac{3}{100}=0,03\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,06\left(mol\right)\\n_{CO_2}=0,03\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,06\cdot36,5}{10\%}=21,9\left(g\right)\\V_{CO_2}=0,03\cdot22,4=0,672\left(l\right)\end{matrix}\right.\)
a) \(n_{MgCO_3}=\dfrac{16,8}{84}=0,2\left(mol\right)\)
PTHH: MgCO3 + 2HCl ---> MgCl2 + CO2 + H2O
0,2--------------------->0,2----->0,2
=> \(m=m_{MgCl_2}=0,2.95=19\left(g\right)\)
\(m_{\text{dd}.sau.p\text{ư}}=150+16,8-0,2.44=158\left(g\right)\)
=> \(C\%_{MgCl_2}=\dfrac{19}{158}.100\%=12,025\%\)
b) CO2 + Ca(OH)2 ---> CaCO3 + H2O
0,2----->0,2------------>0,2
=> \(\left\{{}\begin{matrix}m_{kt}=m_{CaCO_3}=0,2.100=20\left(g\right)\\V_{\text{dd}Ca\left(OH\right)_2}=\dfrac{0,2}{3}=\dfrac{1}{15}\left(l\right)\end{matrix}\right.\)
a)\(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\)
PTHH: Fe + 2HCl → FeCl2 + H2
Mol: 0,25 0,5 0,25
b) \(C_{M_{ddHCl}}=\dfrac{0,5}{0,5}=1M\)
c) \(V_{H_2}=0,25.22,4=5,6\left(l\right)\)
a) Fe + 2HCl --> FeCl2 + H2
b) \(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
_____0,4--->0,8------>0,4--->0,4
=> VH2 = 0,4.22,4 = 8,96(l)
c) mHCl = 0,8.36,5 = 29,2 (g)
=> \(m_{dd\left(HCl\right)}=\dfrac{29,2.100}{7,3}=400\left(g\right)\)
mdd (sau pư) = 22,4 + 400 - 0,4.2 = 421,6 (g)
=> \(C\%\left(FeCl_2\right)=\dfrac{127.0,4}{421,6}.100\%=12,05\%\)
a) $n_{CaCO_3} = 0,15(mol)$
$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
$n_{HCl} = 2n_{CaCO_3} = 0,3(mol)$
$m_{dd\ HCl} = \dfrac{0,3.36,5}{7,3\%} = 150(gam)$
b)
$n_{CaCl_2} = n_{CO_2} = n_{CaCO_3} =0,15(mol)$
$V_{CO_2} = 0,15.22,4 = 3,36(lít)$
c)
$m_{dd} = 15 + 150 - 0,15.44 = 158,4(gam)$
$C\%_{CaCl_2} = \dfrac{0,15.111}{158,4}.100\% = 10,51\%$
\(CaCO_3\left(0,5\right)+2HCl\left(1\right)\rightarrow CaCl_2+H_2O+CO_2\left(0,5\right)\)
\(V_{ddHCl}=\dfrac{1}{0,5}=2\left(l\right)\)
\(\dfrac{n_{KOH}}{n_{CO_2}}=\dfrac{1}{0,5}=2\)
=> xảy ra pư tạo muối TH
\(CO_2\left(0,5\right)+2KOH\rightarrow K_2CO_3\left(0,5\right)+H_2O\)
\(\Rightarrow C_{MddK_2CO_3}=\dfrac{0,5}{0,5}=1M.\)