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a) $CaSO_3 + 2HCl \to CaCl_2 + SO_2 + H_2O$
b)
$n_{SO_2} = n_{CaSO_3} = \dfrac{12}{120} = 0,1(mol)$
$m_{SO_2} = 0,1.64 = 6,4(gam)$
c)
$n_{HCl} = 2n_{SO_2} = 0,2(mol)$
$\Rightarrow m_{dd\ HCl} = \dfrac{0,2.36,5}{14,6\%} = 50(gam)$
d)
$m_{dd\ sau\ pư} = m_{CaSO_3} + m_{dd\ HCl} - m_{SO_2} = 12 + 50 - 6,4 = 55,6(gam)$
$C\%_{CaCl_2} = \dfrac{0,1.111}{55,6}.100\% = 19,96\%$
Ta có: \(n_{CaSO_3}=\dfrac{12}{120}=0,1\left(mol\right)\)
a. PTHH: CaSO3 + 2HCl ---> CaCl2 + H2O + SO2
b. Theo PT: \(n_{SO_2}=n_{CaSO_3}=0,1\left(mol\right)\)
=> \(m_{SO_2}=0,1.64=6,4\left(g\right)\)
c. Theo PT: \(n_{HCl}=2.n_{CaSO_3}=2.0,1=0,2\left(mol\right)\)
=> \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
Ta có: \(C_{\%_{HCl}}=\dfrac{7,3}{m_{dd_{HCl}}}.100\%=14,6\%\)
=> \(m_{dd_{HCl}}=50\left(g\right)\)
d. Ta có: \(m_{dd_{CaCl_2}}=12+50-0,1.64=55,6\left(g\right)\)
Theo PT: \(n_{CaCl_2}=n_{SO_2}=0,1\left(mol\right)\)
=> \(m_{CaCl_2}=0,1.111=11,1\left(g\right)\)
=> \(C_{\%_{CaCl_2}}=\dfrac{11,1}{55,6}.100\%=19,96\%\)
\(a,PTHH:Fe+2HCl\to FeCl_2+H_2\\ b,n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ \Rightarrow n_{Fe}=0,15(mol)\\ \Rightarrow m_{Fe}=0,15.56=8,4(g)\\ c,n_{HCl}=2n_{H_2}=0,3(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,3.36,5}{36,5\%}=30(g)\)
a) PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
b) Ta có: \(n_{Fe}=\dfrac{0,56}{56}=0,01\left(mol\right)=n_{FeSO_4}=n_{H_2}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,01\cdot152=1,52\left(g\right)\\V_{H_2}=0,01\cdot22,4=0,224\left(l\right)\end{matrix}\right.\)
c) Theo PTHH: \(n_{Fe}=n_{H_2SO_4}=0,01\left(mol\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,01\cdot98}{19,6\%}=5\left(g\right)\)
Ta có: \(n_{Fe}=\dfrac{2,24}{56}=0,04\left(mol\right)\)
a. PTHH: Fe + 2HCl ---> FeCl2 + H2
b. Theo PT: \(n_{FeCl_2}=n_{H_2}=n_{Fe}=0,04\left(mol\right)\)
=> \(m_{FeCl_2}=0,04.127=5,08\left(g\right)\)
=> \(V_{H_2}=0,04.22,4=0,896\left(lít\right)\)
c. Theo PT: \(n_{HCl}=2.n_{Fe}=2.0,04=0,08\left(mol\right)\)
=> \(m_{HCl}=0,08.36,5=2,92\left(g\right)\)
Ta có: \(C_{\%_{HCl}}=\dfrac{2,92}{m_{dd_{HCl}}}.100\%=5\%\)
=> \(m_{dd_{HCl}}=58,4\left(g\right)\)