Question

"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?"

Answer 1

Answer:

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

Answer 2

Answer:

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