Question

net ionic equation for SrBr2 + K2SO4 --->

Answer 1

Answer:

$2K^+\text{ + 2Br}^-\text{ }\rightarrow\text{ 2KBr}_{(s)}$

Explanation:

Here, we want to write the net ionic equation for the given equation;

$SrBr_2\text{ + K}_2SO_4\text{ }\rightarrow\text{ SrSO}_4\text{ + 2KBr}$

Now, what is left is to write the participating ions:

$Sr^{2+}\text{ + 2Br}^-\text{ + 2K}^+\text{ + SO}_4^{2-}\text{ }\rightarrow\text{ Sr}^{2+}\text{ + SO}_4^{2-}\text{ + 2KBr}$

We must identify that the Potassium Bromide is the precipitate and thus would not be dissociated into ions

Thus, we have it that:

The Strontium ion cancels out, including the sulphate

We have left:

$2K^+\text{ + 2Br}^-\text{ }\rightarrow\text{ 2KBr}$