Question

A chemist wants to produce 12.00 grams of barium sulfate by reacting a .6000 M BaCl2 solution with excess H2SO as show in the reaction below. What volume of the barium chloride should be used?
BaCl2 + H2SO4 --> BaSO4 +2HCl

Answer 1

Find the moles of BaSO4 first. Then since we know it's a one to one ratio from barium chloride to barium sulfate we can just solve for liters.
First you need to find the moles BaSO4 , and the you will require to find barium sulfate in liters.

12.00gBaSO4 / 233.31 grams per mole

=.05141moles

Molarity=moles/liters

Hence,

Liters=.05141moles/.6Molarity
=.85 liters

Answer 2

Answer: 0.0857 L $BaCl_2$

Explanation: It's a stoichiometry problem. Balanced equation is given from which there is 1:1 mol ratio between $BaCl_2$ and $BaSO_4$ .

Chemist wants to produce 12.00 grams of barium sulfate by reacting a 0.6000 M barium chloride solution with excess sulfuric acid.

We know that molarity is moles of solute per liter of solution. Molarity of barium chloride is given. If we know its moles then its volume could easily be calculated.

From given grams of barium sulfate, we calculate its moles and then using mol ratio we calculate the moles of barium chloride.

Molar mass of barium sulfate is 233.38 gram per mol.

The complete set up is shown below using dimensional analysis:

$12.00gBaSO_4(\frac{1molBaSO_4}{233.38gBaSO_4})(\frac{1molBaCl_2}{1molBaSO_4})(\frac{1LBaCl_2}{0.6000molBaCl_2})$

= 0.0857 L $BaCl_2$

So, 0.0857 L or 85.7 mL of $BaCl_2$ should be used.