gọi x la so mol cua HCl
y la so mol cua HNO3
nCaO= \(\dfrac{11,2}{56}=0,2\left(mol\right)\)
CaO + 2HCl \(\rightarrow\) CaCl2 + H2O (1)
de: \(\dfrac{x}{2}\) \(\leftarrow\) x \(\rightarrow\) \(\dfrac{x}{2}\rightarrow\) \(\dfrac{x}{2}\)
CaO + 2HNO3 \(\rightarrow\) Ca(NO3)2 + H2O (2)
de: \(\dfrac{y}{2}\leftarrow\) y \(\rightarrow\dfrac{y}{2}\) \(\rightarrow\dfrac{y}{2}\)
Ta co: CM HCl = \(\dfrac{n_{HCl}}{V}=0,2M\)
CM HNO3 = \(\dfrac{n_{HNO_3}}{V}=0,2M\)
\(\Rightarrow\dfrac{n_{HCl}}{V}=\dfrac{n_{HNO_3}}{V}\Rightarrow n_{HCl}=n_{HNO_3}\)
\(\Rightarrow x=y\)
Ta co: \(\dfrac{x}{2}+\dfrac{y}{2}=0,2\Rightarrow\dfrac{2x}{2}=0,2\)
\(\Rightarrow x=y=0,2\left(mol\right)\)
Ta co: \(\Rightarrow\dfrac{n_{HCl}}{V}+\dfrac{n_{HNO_3}}{V}=0,4\)
\(\Rightarrow V=\dfrac{2x}{0,4}=1l\)
\(m_{CaCl_2}=0,2.111=22,2g\)
\(m_{Ca\left(NO_3\right)_2}=0,2.160=32g\)
PTHH :
CaO + 2HCl -> CaCl2 + H2O (1)
x.............2x............x
CaO + 2HNO3 -> Ca(NO3)2 + H2O (2)
y...............2y...............y
nCaO = 11,2/56=0,2(mol)
Gọi nCaO ở(1) và (2 ) lần lượt là x,y(x,y>0)
=> x+y=0,2
Vì VHCl + VHNO3 = V
<=> V=\(\dfrac{n_{HCl}}{C_{M\left(HCl\right)}}+\dfrac{n_{HNO3}}{C_{M\left(HNO3\right)}}=\dfrac{2x}{0,2}+\dfrac{2y}{0,2}=10\left(x+y\right)\)
= 10.0,2 = 2(l)
============ mới nghĩ ra đến đây thôi , cái mmuối mai làm tiếp
