Na2SO4 + Ba(NO3)2 ----> 2 NaNo3 + BaSO4
moles of sodium sulfate = 0.04*0.1 = 4*10^-3 moles
moles of Barium nitrate = .03*0.2 = 6*10^-3 moles
so the limiting reagent is Na2SO4
so moles of BaSO4 formed = 4 millimoles
mass = 4*10^-3 * 233 = 0.932 grams
find the mass of barium sulfate formed when 30. mL of 0.2 M barium nitrate is...
What mass of barium sulfate (233 g/mol) is produced when 125 mL of a 0.150 M solution of barium chloride is mixed with 25 mL of a 0.175 M solution of iron(III) sulfate
date ory Assignment (d) How many moles of barium sulfate were ac- tually formed? mol answer (a) What was the percent yield of barium sulfate? (d) How many mo tually formed? 1. Read in an appropriate reference a discussion of precipitation, filtration, and drying of precipitates and of mass measurements. 2. Barium sulfate, Baso was prepared by reacting 100.0 mL of 0.0500M potassium sulfate solution with 100.0 mL of 0.0700M barium nitrate solution, follow- ing the procedure of this experiment....
Suppose 1.55 g of barium nitrate is dissolved in 250. mL of a 52.0 m M aqueous solution of ammonium sulfate. Calculate the final molarity of nitrate anion in the solution. You can assume the volume of the solution doesn't change when the barium nitrate is dissolved in it Round your answer to 3 significant digits.
When 1.50 mL of 3.5-M Barium chloride is mixed with 2.50 mL of 0.65-M sodium sulfate, a precipitate forms. a. Write balanced chemical, ionic and net ionic equations for the reaction, including phase labels. b. Calculate the theoretical yield (in grams) of the precipitate and identify the limiting reactant. c.Calculate the grams of the reactant in excess (left over) after the reaction is complete.
How many grams of barium sulfate are formed from reaction of 10.0 mL of 0.250 M (NH4)2SO4 And excess BaCl2? How many grams of barium sulfate are formed from reaction of 10.0 mL of 0.250 M (NH4)2SO4 and excess BaCl2?
5. If 75.0 mL of a 0.20 M solution of sodium nitrate (NaNO3) is mixed with 25.0 mL of 0.10 M barium nitrate (Ba(NO3)2), what is the molar concentration of nitrate in the resulting solution? A. 0.10 M D. 0.15 M B. 0.20 M E. 0.18 M C. 0.30 M
how many grams of silver chromate will precupitat when 400.mL of 0.200 M silver nitrate are added to 200. mL of 0.600 M lithium chromate?? Leanne Lara Tuesday Stoichiometry Worksheet Solve the following solutions Stoichiometry problems: A AY NO 1. How many grams of silver chromate will precipitate when 400 mL of 0.200 M silver nitrate are added to 200. mL of 0.600 M lithium chromate? ucro 2 How many mL of 0.380 M barium nitrate are required to precipitate...
The molar solubility of barium sulfate in a 0.248 M barium nitrate solution is M.
3. For the reaction between aqueous barium nitrate and aqueous sodium sulfate producing the precipitate (solid) barium sulfate and aqueous sodium nitrate (a) Write an equation for this reaction. (b) Balance the equation from Part (a). (c) Classify the reaction.
A 15.0 mL sample of a 1.60 M potassium sulfate solution is mixed with 14.4 mL of a 0.890 M barium nitrate solution and this precipitation reaction occurs: K2SO4(aq)+Ba(NO3)2(aq)→BaSO4(s)+2KNO3(aq) The solid BaSO4 is collected, dried, and found to have a mass of 2.52 g . Determine the limiting reactant, the theoretical yield, and the percent yield.