Question

What is the balanced equation for the reaction of a solution of sodium sulfate is mixed with strontium chloride?

Answer 1

NaSO⁴(ads) ,ganadicate

Answer 2

`Na2SO_(4(aq))`The balanced equation for the reaction of a solution of sodium sulfate is mixed with strontium chloride is:

`Na_(2) SO_(4) _((aq))`
`+SrCl_{2_((aq))`
`\to 2NaCl_((aq)) +SrSO_(4)_((s))`

This is a precipitation reaction, which means that one of the products is a solid that precipitates out of solution. In this case, the solid product is strontium sulfate,
`SrSO_(4)`.

The balanced equation for the reaction of a solution of sodium sulfate mixed with strontium chloride is :
`Na2SO_(4(aq)) +SrCl_(2(aq)) \to 2NaCl_((aq)) +SrSO_(4(s))` . This is a precipitation reaction, which means that one of the products is a solid that precipitates out of solution. In this case, the solid product is strontium sulfate,
`SrSO_(4)`

When a solution of sodium sulfate,
`Na2SO_(4(aq))`, is mixed with a solution of strontium chloride,
`SrCl_(2(aq))`, a double displacement reaction occurs. In a double displacement reaction, the positive and negative ions of the two reactants switch partners to form two new products.

In this case, the sodium ions, Na+, from the sodium sulfate react with the chloride ions, Cl-, from the strontium chloride to form sodium chloride, NaCl. The strontium ions,
`Sr_(2)`+, from the strontium chloride react with the sulfate ions,
`SO_(42)`-, from the sodium sulfate to form strontium sulfate,
`SrSO_(4)`.

Strontium sulfate is a white, insoluble solid. This means that it will precipitate out of solution and form a white solid at the bottom of the reaction vessel.

This reaction can be used to identify the presence of strontium ions in a solution. If a solution of sodium sulfate is added to a solution of strontium ions and a white precipitate forms, then this indicates the presence of strontium ions in the solution.

Therefore, The balanced equation for the reaction is:

`Na2SO_(4(aq)) +SrCl_(2(aq)) \to 2NaCl_((aq)) +SrSO_(4(s))`