How many millliters of 2.75 M $H_{2} S O_{4} (a q)$ $H_2SO_4(aq)$ are needed to react completely with 47.5 g of $B a O_{2} (s)$ $BaO_2(s)$ ?

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Answer 1 · 2016-07-07T12:56:40

$102 \ mL$

Write a balanced chemical equation for the reaction between barium peroxide and sulfuric acid.

$BaO_2 + H_2SO_4 -> BaSO_4 + H_2O_2$

Use the stoichiometry of the equation to find the number of moles of sulfuric acid reacted.

$47.5 \ g \ BaO_2xx (1 mol. BaO_2)/(169.33 \ g \ BaO_2)xx (1\ mol. H_2SO_4)/(1\ mol.\ BaO_2)$

$47.5 cancel(\ g \ BaO_2)xx (1 color(red)cancel(\ mol. BaO_2))/(169.33 cancel(\ g \ BaO_2))xx (1\ mol. H_2SO_4)/(1color(red)cancel(\ mol.\ BaO_2))$

$0.281 \ mol.\ H_2SO_4$

Now, use the molarity formula to find the volume of the sulfuric acid used.

$C_M = n/ V$

Where:

$C_M " "$ is the molarity of the acid solution expressed in mol/L.

$V " "$ is the volume of the acid solution expressed L.

$n" "$ is the number of moles.

$V = n / C_M$

$V = ( 0.281 \ mol.)/(2.75 \ mol. L^-1)$

$V = 0.102 \ L$

$V = 102 \ mL$

How many millliters of 2.75 M H_2SO_4(aq)H2SO4(aq)H_2SO_4(aq) are needed to react completely with 47.5 g of BaO_2(s)BaO2(s)BaO_2(s)?