Excess Na2SO4 (aq) is added to a 42.53 mL sample of Ba(NO3)2(aq).

1. What is the formula of the precipitate?
The formula for the precipitate is 2Na(NO3)(aq).

2. If 3.046 grams of precipitate was formed, what was the molarity of the Ba(NO3)2(aq)?

Answers

Answer 1

1. The formula of the precipitate formed is Ba₂SO₄(s)

2. The molarity of the Ba(NO₃)₂ solution is 0.188 M

1. Determination of the formula of the precipitate formed.

We'll begin by writing the net ionic equation for the reaction between Na₂SO₄ and Ba(NO₃)₂. This is given below:

Na₂SO₄(aq) + Ba(NO₃)₂(aq) —>

2Na⁺(aq) + SO₄²¯(aq) + Ba²⁺(aq) + 2NO₃¯(aq) —> Ba₂SO₄(s) + 2Na⁺(aq) + 2NO₃¯(aq)

Cancel the spectator ions (i.e Na⁺ and NO₃¯) to obtain the net ionic equation

SO₄²¯(aq) + Ba²⁺(aq) —> Ba₂SO₄(s)

Therefore, the formula of the precipitate formed is Ba₂SO₄(s)

2. Determination of the molarity of Ba(NO₃)₂

We'll begin by calculating the number of mole of in 3.046 g of precipitate, Ba₂SO₄

Molar mass of Ba₂SO₄ = (2×137) + 32 + (4×16) = 370 g/mol

Mass of Ba₂SO₄ = 3.046 g

Mole of Ba₂SO₄ =?

Mole = mass /molar mass

Mole of Ba₂SO₄ = 3.046 / 370

Mole of Ba₂SO₄ = 0.008 mole

Next, we shall determine the number of mole of Ba(NO₃)₂ required to produce 0.008 mole of Ba₂SO₄

Na₂SO₄(aq) + Ba(NO₃)₂(aq) —> Ba₂SO₄(s) + 2NaNO₃(aq)

From the balanced equation above,

1 mole of Ba(NO₃)₂ reacted to produce 1 mole Ba₂SO₄.

Therefore,

0.008 mole of Ba(NO₃)₂ will also react to produce 0.008 mole of Ba₂SO₄

Thus, 0.008 mole of Ba(NO₃)₂ is required for the reaction.

Finally, we shall determine the molarity of the Ba(NO₃)₂ solution

Mole of Ba(NO₃)₂ = 0.008 mole

Volume = 42.53 mL = 42.53 / 1000 = 0.04253 L

Molarity of Ba(NO₃)₂ =?

Molarity = mole / Volume

Molarity of Ba(NO₃)₂ = 0.008 / 0.04253

Molarity of Ba(NO₃)₂ = 0.188 M

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Related Questions

calculate the mass of butane needed to produce 64.1 g of carbon dioxide to three significant figures and appropriate units

Answers

Answer:

21.2 gm

Explanation:

calculate the mass of butane needed to produce 64.1 g of carbon dioxide to three significant figures and appropriate units

butane is the hydrocarbon C4H10  

in combustion, we react hydrocarbons with O2 to form CO2 and H2O

so

C4H10  + O2---------------->  CO2 + H2O

BALANCE

2C4H10 + 1302--------> 8CO2 + 10 H2O

the molar mass of CO2 is 12 + 16X2 = 44

64.1 gm of CO2 is

64.1/44 = 1.46 MOLES OF  CO2,

FOR EVERY 8 MOLES OF CO2 WE NEED 2 MOLES OF BUTANE  IT IS A

8:2 OR 4:1 RATIO.  THE MOLES OF C4H10 ARE 1/4 THE MOLES OF CO2

SO

THE MOLES OF C4H10 H10 ARE 1.46/4 =0.365 MOLES

THE MOLAR MASS OF BUTANE IS 58.12

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Answer:

Hey mate.....

Explanation:

This is ur answer.....

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Answers

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Explanation:

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