Calculate the mass of the precipitate that precipitates when an excess of potassium carbonate interacts with 170 g

Calculate the mass of the precipitate that precipitates when an excess of potassium carbonate interacts with 170 g of a solution of barium nitrate with a mass fraction of the latter of 16%.

1. Let’s write down the reaction equation:

K2CO3 + Ba (NO3) 2 = 2KNO3 + BaCO3 ↓.

2. Find the mass of barium nitrate and convert it to the amount:

m (Ba (NO3) 2) = (m (solution) * ω (Ba (NO3) 2)) / 100% = (170 g * 16%) / 100% = 27.2 g.

n (Ba (NO3) 2) = m (Ba (NO3) 2) / M (Ba (NO3) 2) = 27.2 g / 261 g / mol = 0.104 mol.

3. Using the equation, we find the amount of sediment, and then its mass:

n (BaCO3) = n (Ba (NO3) 2) = 0.104 mol.

m (BaCO3) = n (BaCO3) * M (BaCO3) = 0.104 mol * 197 g / mol = 20.5 g.

Answer: m (BaCO3) = 20.5 g.