Respuesta :
Ba(OH)₂ + 2HCl = BaCl₂ + 2H₂O
n(HCl)=c(HCl)v(HCl)
n{Ba(OH)₂)}=c{Ba(OH)₂}v{Ba(OH)₂}=2n(HCl)=2c(HCl)v(HCl)
v{Ba(OH)₂}=2c(HCl)v(HCl)/c{Ba(OH)₂}
v{Ba(OH)₂}=2*0.250*0.020/0.100=0.100 L = 100 mL
n(HCl)=c(HCl)v(HCl)
n{Ba(OH)₂)}=c{Ba(OH)₂}v{Ba(OH)₂}=2n(HCl)=2c(HCl)v(HCl)
v{Ba(OH)₂}=2c(HCl)v(HCl)/c{Ba(OH)₂}
v{Ba(OH)₂}=2*0.250*0.020/0.100=0.100 L = 100 mL
Answer:
25.0 mL of 0.100 M Ba(OH)₂ are required.
Explanation:
The reaction that takes place is:
Ba(OH)₂ + 2HCl → BaCl₂ + 2H₂O
The formula for concentration is:
C = n / V
- With the volume and concentration of the HCl solution, we can calculate the moles of HCl, keeping in mind that 20.0 mL = 0.020 L:
0.020 L * 0.250 M = 0.005 mol HCl
Then we convert them to moles of Ba(OH)₂:
0.005 mol HCl * [tex]\frac{1molBa(OH)_{2}}{2molHCl}[/tex] = 0.0025 mol Ba(OH)₂
- Finally, with the moles of Ba(OH)₂ and the concentration we can calculate the volume:
0.0025 mol Ba(OH)₂ / 0.100 M = 0.025 L = 25 mL