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Blizzard [7]
3 years ago
14

Which net ionic equation best represents the reaction that occurs when an aqueous solution of barium chloride is mixed with an a

queous solution of lithium sulfate?
Chemistry
1 answer:
vovangra [49]3 years ago
6 0

Answer : The net ionic equation will be:

Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)

Explanation :

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The balanced molecular equation will be,

BaCl_2(aq)+Li_2SO_4(aq)\rightarrow 2LiCl(aq)+BaSO_4(s)

The complete ionic equation in separated aqueous solution will be,

Ba^{2+}(aq)+2Cl^-(aq)+2Li^+(aq)+SO_4^{2-}(aq)\rightarrow 2Li^+(aq)+2Cl^-(aq)+BaSO_4(s)

In this equation the species, Li^+\text{ and }Cl^{-} are the spectator ions.

By removing the spectator ions , we get the net ionic equation.

The net ionic equation will be:

Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)

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Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica
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<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = \frac{1}{2}\times 0.52=0.26mol of lead (II) iodide

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

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