"11. Barium nitrate reacts with aqueous sodium sulfate to produce solid barium sulfate and aqueous sodium nitrate. Abigail places 20.00 mL of 0.500 M barium nitrate in a flask. She has a 0.225M sodium sulfate solution available. What volume of this solution must she add to her flask of barium nitrate so she has no excess reactant left over?"

44 mL of Na2SO4

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

Ba(NO3)2 (aq) + Na2SO4 (aq) —> BaSO4 (s) + 2NaNO3 (aq)

Step 2:

Determination of the number of mole of Ba(NO3)2 in 20.00 mL of 0.500 M barium nitrate (Ba(NO3)2). This is illustrated below:

Molarity of Ba(NO3)2 = 0.5 M

Volume of solution = 20 mL = 20/1000 = 0.02 L

Mole of solute (Ba(NO3)2) =?

Molarity = mole /Volume

0.5 = Mole of Ba(NO3)2 / 0.02

Cross multiply to express in linear form

Mole of Ba(NO3)2 = 0.5 x 0.02

Mole of Ba(NO3)2 = 0.01 mole

Step 3:

Determination of the number of mole of Na2SO4 that reacted.

Ba(NO3)2 (aq) + Na2SO4 (aq) —> BaSO4 (s) + 2NaNO3 (aq)

From the balanced equation above,

1 mole of Ba(NO3)2 reacted with 1 mole of Na2SO4.

Therefore, 0.01 mole of Ba(NO3)2 will also react with 0.01 mole of Na2SO4.

Step 4:

Determination of the volume of Na2SO4 needed for the reaction. This is illustrated below:

Mole of Na2SO4 = 0.01 mole

Molarity of Na2SO4 = 0.225M

Volume =?

Molarity = mole /Volume

0.225 = 0.01 / volume

Cross multiply to express in linear form

0.225 x Volume = 0.01

Divide both side by 0.225

Volume = 0.01/0.225

Volume of Na2SO4 = 0.044 L

Converting 0.044 L to mL, we have

Volume of Na2SO4 = 0.044 x 1000

Volume of Na2SO4 = 44 mL

Therefore, 44 mL of Na2SO4 is needed for the reaction

She has to add 44.44 mL of Na2SO4

Explanation:

Step 1: Data given

Volume of barium nitrate = 20.00 mL = 0.020 L

Molarity of barium nitrate = 0.500 M

Molarity of sodium sulfate = 0.225 M

Step 2: The balanced equation

Ba(NO3)2 (aq) + Na2SO4 (aq) → BaSO4 (s) + 2 NaNO3 (aq)

Step 3: Calculate volume of the sodium sulfate solution needed

For 1 mol Ba(NO3)2 we need 1 mol Na2SO4 to produce 1 mol basO4 and 2 moles NaNO3

C1*V1 = C2*V2

⇒with C1 = the molarity of Ba(NO3)2 = 0.500 M

⇒with V1 = the volume of Ba(NO3)2 = 0.020 L

⇒with C2 = the molarity of Na2SO4 =0.225 M

⇒with V2 = the volume of Na2SO4 = TO BE DETERMINED

0.500 M * 0.020 L = 0.225 M * V2

V2 = (0.500 M * 0.020 L) / 0.225 M

V2 = 0.04444 L = 44.44 mL

She has to add 44.44 mL of Na2SO4

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