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Anton [14]
3 years ago
7

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.
Chemistry
1 answer:
Vladimir [108]3 years ago
3 0

Answer:

There will be formed 3.03 grams of the precipitate BaSO4

Explanation:

<u>Step 1: </u>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

<u>Step 2:</u> 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.

<u>Step 3</u>: 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

<u>Step 4:</u> Calculate moles Na2SO4

Moles Na2SO4 = 1.30M * 0.01 L

Moles Na2SO4 = 0.013 moles

<u>Step 5</u>: 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

<u>Step 6:</u> 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

<u>Step 7:</u> 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

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