Question

[Ivy Master]

cứu

Answer 1

[Rhider]

C và B

Answer 2

[Nguyễn Hoàng Minh]

Câu 12:

\(m_{KOH}=\dfrac{44,8\cdot25\%}{100\%}=11,2\left(g\right)\\ \Rightarrow n_{KOH}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ PTHH:KOH+HCl\rightarrow KCl+H_2O\\ \Rightarrow n_{HCl}=0,2\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,2\cdot36,5=7,3\left(g\right)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{7,3\cdot100\%}{2,5\%}=292\left(g\right)\\ \text{Chọn }C\)