Question

Answer 1

Câu 37 : 

\(n_{HCl}=1,5.0,05=0,075\left(mol\right)\)

1) PTHH : \(KOH+HCl\rightarrow KCl+H_2O\)

Theo Pt : \(n_{HCl}=n_{KOH}=n_{KCl}=0,075\left(mol\right)\)

2) \(V_{ddKOH}=\dfrac{0,075}{2}=0,0375\left(l\right)\)

3) \(C_{MKCl}=\dfrac{0,075}{0,0375+0,05}=\dfrac{6}{7}\left(M\right)\)

 

Answer 2

\(1.KOH+HCl\rightarrow KCl+H_2O\\ 2.n_{HCl}=0,05.1,5=0,075mol\\ n_{KOH}=n_{KCl}=n_{HCl}=0,075mol\\ V_{ddKOH}=\dfrac{0,075}{2}=0,0375l\\ 3.C_{M_{KCl}}=\dfrac{0,075}{0,0375+0,05}=\dfrac{6}{7}M\)