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

1 Answer
Jul 3, 2017

Answer:

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.