Question

При взаимодействии 60 грамм сульфата натрия с избытком раствора хлорида бария выпал осадок. Расчитайте массу полученного осадка и его количество вещества.

Answer 1

Дано
m(Na2SO4)=60 g
BaCL2
----------------------
mBaSO4-?
n(BaSO4)-?
  60                               Xg
 Na2SO4+BaCL2-->BaSO4+2NaCL
 142                             233
 M(Na2SO4)=142 g/mol, M(BaSO4)=233 g/mol
X=60*233 / 142
X=98.45 g
n(BaSO4)=m/M=98.45/233=0.42 mol
ответ 0.42 моль, 98.45 г

Answer 2

Решение находится на фото

Answer 3

Calculation of Mass and Amount of Precipitate

To calculate the mass and amount of precipitate formed when 60 grams of sodium sulfate reacts with an excess of barium chloride solution, we need to consider the balanced chemical equation for the reaction:

Na2SO4 + BaCl2 -> BaSO4 + 2NaCl

From the equation, we can see that 1 mole of sodium sulfate (Na2SO4) reacts with 1 mole of barium chloride (BaCl2) to produce 1 mole of barium sulfate (BaSO4) and 2 moles of sodium chloride (NaCl).

To calculate the mass of the precipitate, we need to determine the molar mass of barium sulfate (BaSO4). According to the periodic table, the m