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

| August 18, 2021 | education

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.



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