Question

What volume of 5.00 M `H_2SO_4` is required to react with 378 moles Bacl What mass of `BaCl_2`?

Answer 1

Approx. `80*L`............

Explanation:

We need (i), a stoichiometric reaction........

`BaCl_2(aq) + H_2SO_4(aq) rarr BaSO_4(s)darr + 2HCl(aq)`

And (ii), equivalent quantities of sulfuric acid.....

`(378*mol)/(5.0*mol*L^-1)=75.6/(L^-1)=75.6/(1/L)=75.6*L`

There is a mass of `378*molxx208.23*g*mol^-1=78.7*kg` with respect to barium chloride.