When an excess of potassium carbonate solution interacts with a 10% solution

When an excess of potassium carbonate solution interacts with a 10% solution of barium nitrate, the mass of the precipitate formed is 3.94 g. Determine the mass of the barium nitrate solution taken for the experiment.

Given:
w (Ba (NO3) 2) = 10% = 0.1
m (BaCO3) = 3.94 g
To find:
m (Ba (NO3) 2)
Decision:
K2CO3 + Ba (NO3) 2 = 2KNO3 + BaCO3
n (BaCO3) = m / M = 3.94 g / 197 g / mol = 0.02 mol
n (Ba (NO3) 2): n (BaCO3) = 1: 1
n (Ba (NO3) 2) = 0.02 mol
m (Ba (NO3) 2) = n * M = 0.02 mol * 261 g / mol = 5.22 g
Answer: 5.22 g