This is a limiting reagent (LR) problem. One reagent is the LR, the other is the excess reagent (ER). I do these the long way; i.e., there are shorter ways to do them but I have trouble explaining them.
Ba(OH)2 + Na2SO4 ==> BaSO4 + 2NaOH
1. mols Ba(OH)2 = M x L = ?
2. mols Na2SO4 ==> M x L = ?
3. Using the coefficients in the balanced equation, convert mols Ba(OH)2 to mols BaSO4.
4. Do the same to convert mols Na2SO4 to mols BaSO4.
5. You can see that mols BaSO4 formed is not the same for both 1 and 2. In LR problems, the SMALLER number is ALWAYS the correct for the product formed the reagent producing that number is the LR.
6. Convert the smaller number into grams. g BaSO4 = mols BaSO4 x molar mass BaSO4.
Calculate the mass of the precipitate formed when 2.27 L of 0.0820 M Ba(OH)2 are mixed with 3.06 L of 0.0664 M Na2SO4.
3 answers
idk
It appears that you may need more help or clarification. Is there a particular part of the problem or solution that you are unsure about?
2 answers