What is the mass of the precipitate formed during the interaction of a potassium carbonate

What is the mass of the precipitate formed during the interaction of a potassium carbonate solution containing 10.6 g of a substance and silver netrate?

To solve this problem, we write down given: m (K2CO3) = 10.6 g

Find: sediment mass -?

Decision:

Let’s write down the reaction equation.

AgNO3 + K2CO3 = Ag2CO3 + KNO3

In this reaction, the resulting precipitate has the formula Ag2CO3.

Let’s arrange the coefficients.

2AgNO3 + K2CO3 = Ag2CO3 + 2KNO3

Let’s calculate the molar masses of potassium carbonate and silver carbonate.

M (K2CO3) = 39 * 2 + 12 + 16 * 3 = 138 g / mol

M (Ag2CO3) = 108 * 2 + 12 + 16 * 3 = 276 g / mol

We write 10.6 g over potassium carbonate, and 138 g under potassium carbonate.

We write x g above the silver carbonate, and 276 g below the silver carbonate.

Let’s compose and solve the proportion:

x = 10.6 * 276/138 = 21.2 g

Answer: 21.2 g