Question

Given the net ionic equation, Ba2+ + SO42- => BaSO4, how many grams of barium chloride must be present to react with 200 grams of iron(III) sulfate? (MMFe2(SO4)3 = 400 g/mol, MMBaCl2 = 208 g/mol)

Answer 1

Answer: 208g/mol

Explanation:

First of all we have to write the balance equation for the reaction.

BaCl2 + Fe2(SO4)3____> BaSO4 + FeCl3

After balancing we have.

3BaCl2 +Fe2(SO4)3_____> 3BaSO4 +2FeCl3

Looking at the equation, we find out that 3 moles of barium chloride reacts with 1 mole of iron iii sulfate

Therefore we have

3moles of BaCl2 _____> 400g/mole of iron iii sulfate

Xmole of BaCl2 _____> 200g/mole of iron iii sulfate

X = 2 * 200g/mol divide by 400g/mol

X = 1mole

1 mole of BaCl2 will be need to react with 200g/mol of iron iii sulfate.

This 1 mole of BaCl2 is equivalent to 208g/mol of BaCl2.

Therefore the gram of barium chloride that must be present is = 208g/mol//