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

**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.**

Let’s find the amount of substance BaCO3.

n = m: M.

M (BaCO3) = 197 g / mol.

n = 3.94 g: 197 g / mol = 0.02 mol.

Let’s find the quantitative ratios of substances.

К2СО3 + Ba (NO3) 2 = BaCO3 ↓ + 2KNO3

For 1 mol of BaCO3, there is 1 mol of Ba (NO3) 2.

The substances are in quantitative ratios of 1: 1.

The amount of substance will be equal.

n (BaCO3) = n (Ba (NO3) 2) = 0.02 mol.

Let us find the mass of Ba (NO3) 2.

m = n × M.

М (Ba (NO3) 2) = 261 g / mol.

m = 261 g / mol × 0.02 mol = 5.22 g.

Let us find the mass of Ba (NO3) 2 in solution.

W = m (substance): m (solution) × 100%,

m (solution) = m (substance): w) × 100%.

m (solution) = (5.22: 10%) × 100% = 52.2 g.

Answer: 52.2g.