Question

how many milliliter of 2.0M HCl are required to react with 4.40g of solid containing 26.13wt% Ba(NO3)2 if the reaction is

Ba^2+(aq)+ 2Cl-(aq)-->BaCl2(aq)

Answer 1

mass of Ba(NO3)2 = 26.13 % of sample mass

= 26.13 % of 4.40 g

= 26.13*4.40/100

= 1.15 g

Molar mass of Ba(NO3)2,

MM = 1*MM(Ba) + 2*MM(N) + 6*MM(O)

= 1*137.3 + 2*14.01 + 6*16.0

= 261.32 g/mol

mass(Ba(NO3)2)= 1.15 g

number of mol of Ba(NO3)2,

n = mass of Ba(NO3)2/molar mass of Ba(NO3)2

=(1.15 g)/(261.32 g/mol)

= 4.401*10^-3 mol

So,

mol of Ba2+ = 4.401*10^-3 mol

from reaction ,

mol of HCl = 2*mol of Ba2+

= 2*4.401*10^-3 mol

= 8.802*10^-3 mol

now use:

mol(HCl) = M(HCl)*V(HCl)

8.802*10^-3 = 2.0 M * V

V= 4.40*10^-3 mL

V= 4.40 mL

Answer: 4.40 mL