How many milliliters of 3.00 M H2SO4 are required to react with 4.35 g of solid containing 23.2 wt% Ba(NO3)2 if the reaction is Ba21 1 SO22 n BaSO (s)?

1.29 mL.

Explanation:

Equation of the reaction:

H2SO4(aq) + Ba(NO3)2(aq) --> BaSO4 + 2HNO3(aq)

Mass of Ba(NO3)3 = wt% * mass of the solid

= 23.2 x 4.35 / 100

= 1.01 g

Number of moles of Ba(NO3)2 = mass/molar mass

Molar mass of Ba(NO3)2 = 137 + (14 + (16 * 3))*2

= 261 g/mol

= 1.01 / 261

= 0.00387

By stoichiometry, since 1 mole of Ba(NO3)2 reacted with 1 mole of H2SO4. Therefore, number of moles of H2SO4 = 0.00387 moles.

Volume = number of moles/molar concentration

= 0.00387/3

= 0.00129 L

= 1.29 mL.

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~courtney