a,\(NaOH+HCl\rightarrow NaCl+H_2O\)
\(n_{HCl}=\dfrac{m}{M}=\dfrac{20}{36,5}=\dfrac{40}{73}\left(mol\right)\)
Theo PTHH : \(n_{NaCl}=n_{HCl}=\dfrac{40}{73}\left(mol\right)\)
\(\Rightarrow m_{NaCl}=n.M=\dfrac{40}{73}.58,5=32,1\left(g\right)\)
b, \(m_{ddX}=m_{HCl}+m_{ddNaOH}=20+200=220\left(g\right)\)
n HCl = \(\dfrac{m}{M}=\dfrac{20}{36,5}\approx0,5\left(mol\right)\)
a)
Ptr:
NaOH + HCl \(\rightarrow\) NaCl + H2O
1..............1...........1.........1
0,2..........0,2........0,2.......0,2
m NaCl = n.M = 0,2. 58,5= 11,7(g)
Vậy khối lượng NaCl thu được là 11,7g
b)
khối lượng dd x thu được là:
mdd X = m NaOH + m HCl= 200+20=220(g)
