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\(a)n_{Fe}=\dfrac{16,8}{56}=0,3mo\\ n_{H_2SO_4}=\dfrac{196.20}{100.98}=0,4mol\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ \Rightarrow\dfrac{0,3}{1}< \dfrac{0,4}{1}\Rightarrow H_2SO_4.dư\\ n_{Fe}=n_{H_2SO_4}=n_{FeSO_4}=n_{H_2}=0,3mol\\ V_{H_2\left(đkc\right)}=0,3.24,79=7,437l\\ b)C_{\%FeSO_4}=\dfrac{0,2.152}{16,8+196-0,3.2}\cdot100=14,32\%\\ C_{\%H_2SO_4}=\dfrac{\left(0,4-0,3\right).98}{16,8+196-0,3.2}\cdot100=4,62\%\)
\(A,\) Tên khí A là SO2
\(B,\) \(n_{SO_2}=\frac{1.12}{22.4}=0.05\left(mol\right)\)
\(K_2SO_3+H_2SO_4\rightarrow K_2SO_4+SO_2+H_2O\)
0.05 0.05 0.05
\(m_{H_2SO_4}=0.05\times98=4.9\left(g\right)\)
\(C\%_{H_2SO_4}=\frac{4.9}{24.5}\times100=20\%\)
\(C,\)
\(m_{K_2SO_3}=0.05\times158=7.9\left(g\right)\)
\(\%m_{K_2SO_3}=\frac{7.9}{21}\times100=37.6\%\)
\(\%m_{K_2SO_4}=100\%-37.6\%=62.4\%\)
a, \(Na_2SO_3+H_2SO_4\rightarrow Na_2SO_4+SO_2+H_2O\)
\(K_2SO_3+H_2SO_4\rightarrow K_2SO_4+SO_2+H_2O\)
Ta có: \(n_{SO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=n_{SO_2}=0,3\left(mol\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,3.98}{20\%}=147\left(g\right)\)
b, Ta có: 126nNa2SO3 + 158nK2SO3 = 44,2 (1)
Theo PT: \(n_{SO_2}=n_{Na_2SO_3}+n_{K_2SO_3}=0,3\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Na_2SO_3}=0,1\left(mol\right)\\n_{K_2SO_3}=0,2\left(mol\right)\end{matrix}\right.\)
Có: m dd sau pư = 44,2 + 147 - 0,3.64 = 172 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{Na_2SO_3}=\dfrac{0,1.126}{172}.100\%\approx7,33\%\\C\%_{K_2SO_3}=\dfrac{0,2.158}{172}.100\%\approx18,37\%\end{matrix}\right.\)
c, \(n_{Ba\left(OH\right)_2}=0,5.1=0,5\left(mol\right)\)
\(\Rightarrow\dfrac{n_{SO_2}}{n_{Ba\left(OH\right)_2}}=0,6< 1\) → Pư tạo BaSO3.
PT: \(SO_2+Ba\left(OH\right)_2\rightarrow BaSO_3+H_2O\)
\(n_{BaSO_3}=n_{SO_2}=0,3\left(mol\right)\Rightarrow m_{BaSO_3}=0,3.217=65,1\left(g\right)\)
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
$m_{K_2CO_3} = 0,1.138 = 13,8(gam) ; m_{Na_2CO_3} = 0,1.106 = 10,6(gam)$
Bài 2 :
\(n_{CaCO3}=\dfrac{20}{100}=0,2\left(mol\right)\)
\(m_{ct}=\dfrac{20.91,25}{100}=18,25\left(g\right)\)
\(n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)
Pt : \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O|\)
1 2 1 1 1
0,2 0,5 0,2 0,2
Lập tỉ số so sánh : \(\dfrac{0,2}{1}< \dfrac{0,5}{2}\)
⇒ CaCO3 phản ứng hết , HCl dư
⇒ Tính toán dựa vào số mol của CaCO3
\(n_{CO2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
\(V_{CO2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)
\(n_{CaCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
⇒ \(m_{CaCl2}=0,2.111=22,2\left(g\right)\)
\(n_{HCl\left(dư\right)}=0,5-\left(0,2.2\right)=0,1\left(mol\right)\)
⇒ \(m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\)
\(m_{ddspu}=20+91,25-\left(0,2.44\right)=102,45\left(g\right)\)
\(C_{CaCl2}=\dfrac{22,2.100}{102,45}=21,67\)0/0
\(C_{HCl\left(dư\right)}=\dfrac{3,65.100}{102,45}=3,56\)0/0
Chúc bạn học tốt
PTHH: \(K_2SO_3+2HCl\rightarrow2KCl+H_2O+SO_2\uparrow\)
a+b+c) Ta có: \(n_{SO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{KCl}=0,5\left(mol\right)=n_{HCl}\\n_{K_2SO_3}=0,25\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{K_2SO_3}=0,25\cdot158=39,5\left(g\right)=a\\m_{KCl}=0,5\cdot74,5=37,25\left(g\right)\\m_{ddHCl}=\dfrac{0,5\cdot36,5}{10,95\%}\approx166,67\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{SO_2}=0,25\cdot64=16\left(g\right)\)
\(\Rightarrow m_{dd}=m_{K_2SO_3}+m_{ddHCl}-m_{SO_2}=190,17\left(g\right)\) \(\Rightarrow C\%_{KCl}=\dfrac{37,25}{190,17}\cdot100\%\approx19,59\%\)
d) PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)
Theo PTHH: \(n_{NaOH}=n_{HCl}=0,5\left(mol\right)\) \(\Rightarrow V_{NaOH}=\dfrac{0,5}{0,5}=1\left(l\right)\)
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