How many millliters of 275 M $H_2SO_4(aq)$ are needed to react completely with 58.5 g of $BaO_2(s)$ ?

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Answer 1 · 2016-07-07T08:30:58

Are you sure mean barium peroxide, $BaO_2$ , and not barium oxide, $BaO$ .

$BaO_2(s) + H_2SO_4(aq) rarr BaSO_4darr + H_2O_2(aq)$

This used to be used as an industrial method to make hydrogen peroxide.

So $"moles of barium peroxide"$ $=$ $(58.5*g)/(169.33*g*mol^-1)$ $=$ $0.345*mol$ .

We need $1$ $"equiv"$ sulfuric acid.

If $0.275*mol*L^-1$ sulfuric acid is available, we need $(0.345*mol)/(0.275*mol*L^-1)xx1000*mL*L^-1$ $=$ $1255*mL$ .

How many millliters of 275 M H_2SO_4(aq)H_2SO_4(aq) are needed to react completely with 58.5 g of BaO_2(s)BaO_2(s)?