1. Write and balance equation
2. Convert given solution data to moles solute ... moles = Molarity x Volume in Liters
3. Apply moles reactants to balanced equation (divide each mole amount by respective coefficient => Limiting Reagent)
4. Using moles of Limiting reagent, Calculate mole yield of ppt (Barium Sulfate)from equation ratios.
5. Convert moles to grams (moles x formula wt of Barium Sulfate)
Hope this helps.
If 50.00 mL of 0.250 M barium nitrate are combined with 50.00 mL of 0.400 M sodium sulfate, how any grams of precipitate will form?
I'd really appreciate it if someone could walk me through this problem. I don't want the answer, I just want to have a hint at which steps I need to take to solve it.
2 answers
This is a limiting reagent (LR) problem; basically you go through the usual stoichiometry steps twice.
1. Write and balance the equation.
Ba(NO3)2 + Na2SO4 ==> BaSO4(s) + 2NaNO3
2a. mols Ba(NO3)2 = M x L = ?
2b. mols Na2SO4 = M x L = ?
3a. Convert mols Ba(NO3)2 in 2a to mols BaSO4 using the coefficients in the balanced equation. That's ?mols Ba(NO3)2 x [1 mol BaSO4/1 mol Ba(NO3)2]
3b. Do the same and convert mols Na2SO4 to mols BaSO4.
3c. Both answers can't be right in 3a and 3b. The correct answer in LR problems is ALWAYS the smaller number and the reagent responsible for the smaller number is called the LR.
4. Now convert mols BaSO4 to grams BaSO4. grams BaSO4 = mols x molar mass = ?
1. Write and balance the equation.
Ba(NO3)2 + Na2SO4 ==> BaSO4(s) + 2NaNO3
2a. mols Ba(NO3)2 = M x L = ?
2b. mols Na2SO4 = M x L = ?
3a. Convert mols Ba(NO3)2 in 2a to mols BaSO4 using the coefficients in the balanced equation. That's ?mols Ba(NO3)2 x [1 mol BaSO4/1 mol Ba(NO3)2]
3b. Do the same and convert mols Na2SO4 to mols BaSO4.
3c. Both answers can't be right in 3a and 3b. The correct answer in LR problems is ALWAYS the smaller number and the reagent responsible for the smaller number is called the LR.
4. Now convert mols BaSO4 to grams BaSO4. grams BaSO4 = mols x molar mass = ?
1 answer