Question

trộn 50ml dd Ba(OH)₂ 0,05M với 150ml dd HCL 0,1M thu được 200ml dd A. Tính C_m của các chất trong dd A

Answer 1

PTHH: \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{Ba\left(OH\right)_2}=0,05\cdot0,05=0,0025\left(mol\right)\\n_{HCl}=0,15\cdot0,1=0,015\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\frac{0,0025}{1}< \frac{0,015}{2}\) \(\Rightarrow\) HCl còn dư, Ba(OH)₂ phản ứng hết

\(\Rightarrow\left\{{}\begin{matrix}n_{BaCl_2}=0,0025mol\\n_{HCl\left(dư\right)}=0,01mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{BaCl_2}}=\frac{0,0025}{0,2}=0,0125\left(M\right)\\C_{M_{HCl\left(dư\right)}}=\frac{0,01}{0,2}=0,05\left(M\right)\end{matrix}\right.\)

Answer 2

nBa(OH)2=0,05.0,05=0,0025(mol)n

nHCl=0,15.0,1=0,015(mol)

PTHH:Ba(OH)2+2HCl→BaCl2+2H2O

Bđ______0,0025______0,015

Pư______0,0025______0,005____0,0025

Kt________0________0,01______0,0025

CMddHCldư=0,01/0,2=0,05M

CMddBaCl2=0,0025/0.2=0.0125M.