Câu 2:
PTHH: \(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)
Ta có: \(n_{Ca\left(OH\right)_2}=n_{CaO}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(\Rightarrow C\%_{Ca\left(OH\right)_2}=\dfrac{0,1\cdot74}{1000+5,6}\cdot100\%\approx0,74\%\)
\(1.n_{Zn}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{200.7,3}{100.36,5}=0,4\left(mol\right)\)
\(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(bd:\) \(0,1\) \(0,4\)
\(pu:\) \(0,1\) \(0,2\) \(0,1\)
\(spu:\) \(0\) \(0,2\) \(0,1\)
\(\rightarrow m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)
\(2.\\ n_{CaO}=0,1\left(mol\right)\\ PTHH:CaO+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\)
\(\left(mol\right)\) \(0,1\) \(0,1\) \(0,1\)
\(C\%_{ddspu}=\dfrac{0,1.74}{5,6+1000-0,1.2}.100\%=0,736\left(\%\right)\)
Câu 1 :
$n_{Zn} = \dfrac{6,5}{65} = 0,1(mol) ; n_{HCl} = \dfrac{200.7,3\%}{36,5} = 0,4(mol)$
$Zn + 2HCl \to ZnCl_2 + H_2$
Ta thấy :
$n_{Zn} : 1 < n_{HCl} : 2$ nên $HCl$ dư
$n_{ZnCl_2} = n_{Zn} = 0,1 \Rightarrow m_{ZnCl_2} = 0,1.136 = 13,6(gam)$
Câu 2 :
$m_{dd} = 5,6 + 1000 = 1005,6(gam)$
$CaO + H_2O \to Ca(OH)_2$
$n_{Ca(OH)_2} = n_{CaO} = \dfrac{5,6}{56} = 0,1(mol)$
$C\%_{Ca(OH)_2} = \dfrac{0,1.74}{1005,6}.100\% = 0,74\%$
Câu 1:
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\n_{HCl}=\dfrac{200\cdot7,3\%}{36,5}=0,4\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,4}{2}\) \(\Rightarrow\) Kẽm p/ứ hết
\(\Rightarrow n_{ZnCl_2}=0,1\left(mol\right)\) \(\Rightarrow m_{ZnCl_2}=0,1\cdot136=13,6\left(g\right)\)
\(n_{Zn}=0,1\left(mol\right)\)
\(m_{HCl}=\dfrac{200.7,3}{100}=14,6\left(g\right)\)
\(n_{HCl}=0,4\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
BĐ : 0,1 0,4
PƯ: 0,1 0,2 0,1 0,1
SPƯ: 0 0,2 0,1 0,1
\(m_{ZnCl_2}=13,6\left(g\right)\)
