Question

Вычислите массу осадка которая образуется при взаимодействии раствора содержащего 30.5г NaSO4 с раствором BaNO3 взятого в необходимом кол ве

Answer 1

Mr(Na2SO4)=23*2+32+16*4=142   Mr(BaSO4)=137+32+16*4=233

уравнение реакции
 30.5г                            хг
Na2SO4+Ba(NO3)2=BaSO4+2NaNO3
142                           233

30,5г---------------------х г
142г--------------------233г
х=30,5*233/142=50 грамм осадка

Answer 2

Calculating the Precipitate Mass

To calculate the mass of the precipitate formed when a solution containing 30.5g of NaSO4 interacts with a solution of BaNO3, we can use the concept of chemical reactions and stoichiometry.

The balanced chemical equation for the reaction between NaSO4 and BaNO3 is:

2 NaSO4 + Ba(NO3)2 → 2 NaNO3 + BaSO4

From the balanced equation, we can see that 1 mole of NaSO4 reacts with 1 mole of Ba(NO3)2 to produce 1 mole of BaSO4.

Steps to Calculate the Precipitate Mass

1. Calculate the moles of NaSO4: - Moles = Mass / Molar mass - Molar mass of NaSO4 = 22.99 (Na) + 32.06 (S) + 4 * 16.00 (O) = 142.04 g/mol - Moles of NaSO4 = 30.5g / 142.04 g/mol

2. Use the mole ratio from the balanced equation: - From the balanced equation, we know that 2 moles of NaSO4 react with 1 mole of BaSO4.

3. Calculate the moles of BaSO4: - Since the mole ratio is 2:1, the moles of BaSO4 formed will be half the moles of NaSO4.

4. Calculate the mass of BaSO4: - Mass = Moles * Molar mass - Molar mass of BaSO4 = 137.33 (Ba) + 32.06 (S) + 4 * 16.00 (O) = 233.39 g/mol - Mass of BaSO4 = (Moles of BaSO4) * 233.39 g/mol

Calculation

- Moles of NaSO4 = 30.5g / 142.04 g/mol = 0.2149 moles - Moles of BaSO4 = 0.2149 moles / 2 = 0.1075 moles - Mass of BaSO4 = 0.1075 moles * 233.39 g/mol = 25.10g

The mass of the precipitate (BaSO4) formed is approximately 25.10 grams.