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) ?

Question

Ada Burton

Chemistry

25 October, 17:22

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) ?

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Answers (1)

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Santos Mccormick · Answer 1

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.