How many grams of precipitate is formed when 12 mg of sodium sulfate solution interacts with the required amount of barium nitrate?

univerkov | December 8, 2020 | education

Na2SO4 + Ba (NO3) 2 = BaSO4 + 2NaNO3
M (Na2SO4) = 142 grams / mol
M (BaSO4) = 233 gram / mol
let’s make a proportion
142 – 233
12 – x
x = 19.7 grams

One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.