Complete and balance the molecular equation, including the phases, for the reaction of aqueous potassium sulfate, k2so4, and aqueous strontium iodide, sri2.

The reaction between K₂SO₄(aq) and SrI₂(aq) produces KI(aq) and SrSO₄(s) as products. 

The reaction is 
K₂SO₄(aq) + SrI₂(aq) → KI(aq)+ SrSO₄(s) 

To balance the equation both side of the reaction should have same number of atoms in each element.

Right hand side of the reaction has 1 K, 1 I, 1 Sr, 1 S and 4 O atoms while 2 K, 2 I, 1 Sr,1 S and 4 O present in left hand side of the reaction. 
Hence, number of I atoms and number of K atoms are not balanced. 
To balance the K atoms we should add 2 before KI. Then I atoms will be 2 at the right hand side. 

Hence, the balanced reaction equation is
K₂SO₄(aq) + SrI₂(aq) → 2KI(aq)+ SrSO₄(s) 

m=9

step-by-step explanation:

7m+8=71 subtract   8 from 71 which is 63 than divide it by 7m   equal 9.