A solution of sodium sulfate was added to a solution of barium nitrate weighing 20 g

| February 10, 2021 | education

A solution of sodium sulfate was added to a solution of barium nitrate weighing 20 g, calculate the mass of the precipitate formed as a result of the reaction.

To solve this problem, we write down the given: m (Ba (NO3) 2) = 20 g added a solution of sodium sulfate.
Find: the mass of the sediment formed.
Decision:
Let us write down the reaction equation and determine what substance precipitated.
Ba (NO3) 2 + Na2SO4 = BaSO4 + NaNO3
The precipitate that formed as a result of the reaction has the formula BaSO4
Let’s arrange the coefficients
Ba (NO3) 2 + Na2SO4 = BaSO4 + 2NaNO3
Let’s calculate the molar masses of barium nitrate and barium sulfate.
M (Ba (NO3) 2) = 137 + (14 + 16 * 3) * 2 = 261 g / mol
M (BaSO4) = 137 + 32 + 16 * 4 = 233 g / mol
We write 20 g above the barium nitrate, and 261 g / mol under the barium nitrate.
We write x g over barium sulfate, and 233 g / mol under barium sulfate.
Let’s compose and solve the proportion
x = 20 * 233/261 = 17.85 g
Answer: 17.85 g



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