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

Jul 3, 2017

Approx. $80 \cdot L$............

#### Explanation:

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

$B a C {l}_{2} \left(a q\right) + {H}_{2} S {O}_{4} \left(a q\right) \rightarrow B a S {O}_{4} \left(s\right) \downarrow + 2 H C l \left(a q\right)$

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

$\frac{378 \cdot m o l}{5.0 \cdot m o l \cdot {L}^{-} 1} = \frac{75.6}{{L}^{-} 1} = \frac{75.6}{\frac{1}{L}} = 75.6 \cdot L$

There is a mass of $378 \cdot m o l \times 208.23 \cdot g \cdot m o {l}^{-} 1 = 78.7 \cdot k g$ with respect to barium chloride.