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\(CaO+2HCl\rightarrow CaCl2+H2O\)
\(n_{HCl}=0,5.1=0,5\left(mol\right)\)
Lập tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,5}{2}\Rightarrow\)HCl dư.Tính theo CaO
\(n_{CaCl2}=n_{CaO}=0,2\left(Mol\right)\)
\(m_{CaCl2}=0,2.111=22,2\left(g\right)\)
\(a.HCl+NaOH\rightarrow NaCl+H_2O\)
PỨ trung hoà
\(b,n_{NaOH}=0,1.1=0,1mol\\ n_{NaCl}=n_{NaOH}=n_{HCl}0,1mol\\ m=m_{HCl}=0,1.36,5=3,65g\\ c,m_{NaCl}=0,1.58,5=5,85g\\ d,n_{HCl}=\dfrac{73.10}{100.36,5}=0,2mol\\ \Rightarrow\dfrac{0,1}{1}< \dfrac{0,2}{1}\Rightarrow HCl.dư\\ n_{HCl,pứ}=n_{NaOH}=0,1mol\\ m_{HCl,dư}=\left(0,2-0,1\right).36,5=3,65g\)
\(a)n_{Fe}=\dfrac{5,6}{56}=0,1mol\\
n_{HCl}=0,5.1=0,5mol\\
Fe+2HCl\rightarrow FeCl_2+H_2\\
\Rightarrow\dfrac{0,1}{1}< \dfrac{0,5}{2}\Rightarrow HCl.dư\\
Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1 0,2 0,1 0,1
\(m_{HCl\left(dư\right)}=\left(0,5-0,2\right).36,5=10,95g\\ b)m_{FeCl_2}=0,1.127=12,7g\\ c)V_{H_2}=0,1.22,4=2,24l\\ d)C_{\%HCl\left(dư\right)}=\dfrac{10,95}{200}\cdot100=5,475\%\\ C_{\%HCl\left(pư\right)}=\dfrac{0,2.36,5}{200}\cdot100=3,65\%\)
\(m_{NaOH}=\dfrac{200\cdot8}{100}=16\left(g\right)\Rightarrow n_{NaOH}=\dfrac{16}{40}=0,4mol\)
\(NaOH+HCl\rightarrow NaCl+H_2O\)
0,4 0,4 0,4 0,4
a)\(m_{HCl}=0,4\cdot36,5=14,6\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{14,6}{7,3}\cdot100=200\left(g\right)\)
b)\(m_{NaCl}=0,4\cdot58,5=23,4\left(g\right)\)
\(m_{H_2O}=0,4\cdot18=7,2\left(g\right)\)
\(m_{ddsau}=200+200-7,2=392,8\left(g\right)\)
\(\Rightarrow C\%=\dfrac{23,4}{392,8}\cdot100=5,96\%\)
c) \(n_{SO_2}=\dfrac{6,72}{22,4}=0,3mol\)
\(2NaOH+SO_2\rightarrow Na_2SO_4+H_2O\)
0,4 0,3 0,3 0,3
\(m_{Na_2SO_4}=0,3\cdot142=42,6\left(g\right)\)
a) $Na_2O + 2HCl \to 2NaCl + H_2O$
b) $n_{Na_2O} = \dfrac{6,2}{62} = 0,1(mol)$
$n_{HCl} = 0,5(mol)$
Ta thấy :
$n_{Na_2O} : 1 < n_{HCl} : 2$ nên HCl dư
$n_{HCl\ pư} = 2n_{Na_2O} = 0,2(mol)$
$n_{HCl\ dư} = 0,5 - 0,2 = 0,3(mol)$
$C_{M_{HCl}} = \dfrac{0,3}{0,5} = 0,6M$
$C_{M_{NaCl}} = \dfrac{0,2}{0,5} = 0,4M$
\(n_{CaO}=\dfrac{11,2}{56}=0,2\left(mol\right);n_{HCl}=0,5.1=0,5\left(mol\right)\)
PTHH: \(CaO+2HCl\rightarrow CaCl_2+H_2O\)
Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,5}{2}\) => CaO hết, HCl dư
Theo PTHH: \(n_{CaCl_2}=0,2\left(mol\right)\Rightarrow m_{CaCl_2}=0,2.111=22,2\left(g\right)\)
Ta có: \(\left\{{}\begin{matrix}n_{CaO}=\dfrac{11,2}{56}=0,2\left(mol\right)\\n_{HCl}=0,5.1=0,5\left(mol\right)\end{matrix}\right.\)
PTHH: CaO + 2HCl ---> CaCl2 + H2O
Ban đầu: 0,2 0,5
Pư: 0,2 0,4
Sau pư: 0 0,1 0,2
=> mmuối = 0,2.111 = 22,2 (g)