When barium chloride reacts with potassium sulfate, barium sulfate and potassium chloride arc produced. the balanced equation for this reaction is: bacl2(aq) + k2so4(aq) → baso4(s) + 2kcl(aq) if 2 moles of potassium sulfate react, the reaction consumes of barium chloride. the reaction produces of barium sulfate and of potassium chloride.

Ratio between the reactants is 1:1, so if 2 moles of potassium sulfate react, The reaction consumes 2 moles of barium chloride

The reaction produces 2 moles of barium sulfate ( ratio is 1:1) and 4 moles of potassium chloride. (Ratio with this is 2:4)

Explanation:

This is the reaction:

BaCl₂(aq) + K₂SO₄(aq) → BaSO₄(s) + 2KCl(aq)

Ratio between the reactants is 1:1, so if 2 moles of potassium sulfate react, The reaction consumes 2 moles of barium chloride

The reaction produces 2 moles of barium sulfate ( ratio is 1:1) and 4 moles of potassium chloride. (Ratio with this is 2:4)

The reaction consumes 2.0 moles of barium chloride. The reaction produces 2.0 moles of barium sulfate and 4.0 moles of potassium chloride.

Explanation:

Step 1: Data given

2 moles of potassium sulfate react

Step 2: The balanced equation

BaCl2(aq) + K2SO4(aq) → BaSO4(s) + 2KCl(aq)

Step 3: Calculate moles

For 1 mol BaCl2 we need 1 mol K2SO4 to produce 1 mol Ba2SO4 and 2 moles KCl

For 2 moles K2SO4 we need 2 moles BaCl2 to produce 2 moles BaSO4 and 4 moles KCl

The reaction consumes 2.0 moles of barium chloride. The reaction produces 2.0 moles of barium sulfate and 4.0 moles of potassium chloride.

the volume increases

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b