Question

How many grams of a precipitate are formed after mixing 10.0 mL of a 1.30 M solution of sodium sulfate with 20.0 mL of a 1.10 M solution of barium nitrate? The mixture of the two solutions results in the formation of solid barium sulfate and aqueous sodium nitrate.

Answer 1

There will be formed 3.03 grams of the precipitate BaSO4

Step-by-step explanation:

Step 1: Data given

Volume of a 1.10 M Ba(NO3)2 solution = 20.0 mL = 0.02 L

Volume of a 1.30M Na2SO4 solution = 10.0 mL =0.01 L

Molar mass of BaSO4 = 233.38 g/mol

Step 2: The balanced equation

Ba(NO3)2 + Na2SO4 → BaSO4 + 2 NaNO3 → 2Na+ + 2NO3- + BaSO4

NaNO3 will is soluble, BaSO4 is insoluble, what means it's the precipitate.

Step 3: Calculate moles of Ba(NO3)2

Moles Ba(NO3)2 = Molarity * volume

Moles Ba(NO3)2 = 1.10M *0.02 L

Moles Ba(NO3)2 = 0.022 moles

Step 4: Calculate moles Na2SO4

Moles Na2SO4 = 1.30M * 0.01 L

Moles Na2SO4 = 0.013 moles

Step 5: The limiting reactant

The mole fraction is 1:1. Na2SO4 has the smallest number of moles, so it's the limiting reactant here.

Na2SO4 will completely be consumed (0.013 moles).

Ba(NO3)2 is in excess. There will remain 0.022 - 0.013 = 0.009 moles

Step 6: Calculate moles BaSO4

For 1 moles Na2SO4 consumed, we have 1 mole of BaSO4 produced.

For 0.013 moles of Na2SO4 consumed, we have 0.013 moles of BaSO4

Step 7: Calculate mass of BaSO4

Mass BaSO4 = moles BaSO4 * Molar mass BaSO4

Mass BaSO4 = 0.013 moles * 233.38 g/mol

Mass BaSO4 = 3.03 grams

There will be formed 3.03 grams of the precipitate BaSO4