Question

Cíu cíu !!

Answer 1

\(n_{AgNO_3}=1.0,5=0,5\left(mol\right);n_{HCl}=2.0,3=0,6\left(mol\right)\)

PTHH: AgNO₃ + HCl → AgCl ↓ + HNO₃

Mol:      0,5          0,5                          0,5

Ta có: \(\dfrac{0,5}{1}< \dfrac{0,6}{1}\)⇒ AgNO₃ pứ hết, HCl dư

* V_{dd sau pứ} = 0,5+0,6 = 1,1 (l)

\(\Rightarrow C_{M_{ddNO_3}}=\dfrac{0,5}{1,1}=0,4545M\)

\(C_{M_{ddHCldư}}=\dfrac{0,6-0,5}{1,1}=0,0909M\)

\(m_{ddAgNO_3}=500.1,2=600\left(g\right);m_{ddHCl}=300.1,5=450\left(g\right)\)

* m_{dd sau pứ} = 600+450 = 1050 (g)

\(C\%_{ddHNO_3}=\dfrac{0,5.63.100\%}{1050}=3\%\)

\(C\%_{ddHCl}=\dfrac{\left(0,6-0,5\right).36,5.100\%}{1050}=0,348\%\)

Answer 2

nAgNO3​​=1.0,5=0,5(mol);nHCl​=2.0,3=0,6(mol)

PTHH: AgNO₃ + HCl → AgCl ↓ + HNO₃

Mol:      0,5          0,5                          0,5

Ta có: \(\frac{0 , 5}{1} < \frac{0 , 6}{1}\)⇒ AgNO₃ pứ hết, HCl dư

* V_{dd sau pứ} = 0,5+0,6 = 1,1 (l)

\(\Rightarrow C_{M_{d d N O_{3}}} = \frac{0 , 5}{1 , 1} = 0 , 4545 M\)

\(C_{M_{d d H C l d ư}} = \frac{0 , 6 - 0 , 5}{1 , 1} = 0 , 0909 M\)

\(m_{d d A g N O_{3}} = 500.1 , 2 = 600 \left(\right. g \left.\right) ; m_{d d H C l} = 300.1 , 5 = 450 \left(\right. g \left.\right)\)

* m_{dd sau pứ} = 600+450 = 1050 (g)

\(C \%_{d d H N O_{3}} = \frac{0 , 5.63.100 \%}{1050} = 3 \%\)

\(C \%_{d d H C l} = \frac{\left(\right. 0 , 6 - 0 , 5 \left.\right) . 36 , 5.100 \%}{1050} = 0 , 348 \%\)