Bài 3 :
\(n_{CaO}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Pt : \(CaO+2HCl\rightarrow CaCl_2+H_2O|\)
1 2 1 1
0,2 0,4 0,2
a) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)
\(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(m_{ddHCl}=\dfrac{14,6.100}{20}=73\left(g\right)\)
b) \(n_{CaCl2}=\dfrac{0,4.1}{2}=0,2\left(mol\right)\)
⇒ \(m_{CaCl2}=0,2.111=22,2\left(g\right)\)
\(m_{ddspu}=11,2+73=84,2\left(g\right)\)
\(C_{CaCl2}=\dfrac{22,2.100}{84,2}=26,37\)0/0
Chúc bạn học tốt
\(n_{CaO}=\dfrac{m}{M}=\dfrac{11,2}{56}=0,2mol\)
PTHH \(CaO+2HCl->CaCl_2+H_2O\)
0,2 0,4 mol
\(m_{HCl}=n.M=0,4.36,5=14,6g\)
\(m_{dd}=\dfrac{m_{ct}.100\%}{C\%}=\dfrac{14,6.100\%}{20\%}=73g\)