How many millliters of 275 M $H_{2} S O_{4} (a q)$ $H_2SO_4(aq)$ are needed to react completely with 58.5 g of $B a O_{2} (s)$ $BaO_2(s)$ ?

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How many millliters of 275 M $H_{2} S O_{4} (a q)$ $H_2SO_4(aq)$ are needed to react completely with 58.5 g of $B a O_{2} (s)$ $BaO_2(s)$ ?

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Answer 1 · 2016-07-07T08:30:58

Are you sure mean barium peroxide, $B a O_{2}$ $BaO_2$ , and not barium oxide, $B a O$ $BaO$ .

Explanation:

$B a O_{2} (s) + H_{2} S O_{4} (a q) \to B a S O_{4} ⏐ ↓ + H_{2} O_{2} (a q)$ $BaO_2(s) + H_2SO_4(aq) rarr BaSO_4darr + H_2O_2(aq)$

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

So $moles of barium peroxide$ $"moles of barium peroxide"$ $=$ $=$ $\frac{58.5 \cdot g}{169.33 \cdot g \cdot m o l^{- 1}}$ $(58.5*g)/(169.33*g*mol^-1)$ $=$ $=$ $0.345 \cdot m o l$ $0.345*mol$ .

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

If $0.275 \cdot m o l \cdot L^{- 1}$ $0.275*mol*L^-1$ sulfuric acid is available, we need $\frac{0.345 \cdot m o l}{0.275 \cdot m o l \cdot L^{- 1}} \times 1000 \cdot m L \cdot L^{- 1}$ $(0.345*mol)/(0.275*mol*L^-1)xx1000*mL*L^-1$ $=$ $=$ $1255 \cdot m L$ $1255*mL$ .

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How many millliters of 275 M $H_{2} S O_{4} (a q)$ $H_2SO_4(aq)$ are needed to react completely with 58.5 g of $B a O_{2} (s)$ $BaO_2(s)$ ?

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How many millliters of 275 M H2SO4(aq)H_2SO_4(aq) are needed to react completely with 58.5 g of BaO2(s)BaO_2(s)?

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How many millliters of 275 M $H_{2} S O_{4} (a q)$ $H_2SO_4(aq)$ are needed to react completely with 58.5 g of $B a O_{2} (s)$ $BaO_2(s)$ ?