How many moles of aqueous potassium ions and sulfate ions are formed when 63.7 g of K2SO4 dissolves in water

WHEN 63.7 g OF K2SO4 IS DISSOLVED IN WATER, 0.73 MOLES OF POTASSIUM ION AND 0.366 MOLES OF SULFATE ION ARE FORMED.

Explanation:

Equation for the reaction:

K2SO4 + H20 >2 K+ + SO4^2-

When K2SO4 dissolves in water, potassium ion and sulfate ion are formed.

1 mole of K2SO4 produces 2 moles and 1 mole of SO4^2-

At STP, 1 mole of K2SO4 will be the molar mass of the substance

Molar mass of K2SO4 = ( 39 *2 + 32 + 16*4) g/mol

Molar mass = 174 g/mol

So therefore;

1 mole of K2SO4 contains 174 g and it produces 2 moles of potassium and 1 mole of sulfate ion

When 63.7 g is used; we have:

174 g = 2 moles of K+

63.7 g = ( 63.7 * 2 / 174) moles of K+

= 0.73 moles of K+

Forr sulfate ion, we have:

174 g = 1 mole ofSO4^2-

63.7 g = (63.7 * 1 / 174) moles of SO4^2-

= 0.366 moles of SO4^2-

In other words, when 63.7 g of K2SO4 is dissolved in water, 0.73 moles of potassium ion and 0.366 moles of sulfate ion are formed.

im not exactly sure but i think its b!