1.29 milliliters of 3.00 m [tex]H_{2}[/tex] [tex]SO_{4}[/tex] (Sulfuric acid) are required to react with 4.35 g of a solid mixture containing 23.2 % (w/w) [tex]Ba(NO_{3)2[/tex]
Equation of the reaction:
[tex]H_2[/tex]S[tex]O_{4}[/tex](aq) + Ba(N[tex]O_{3}[/tex][tex])_{2}[/tex](aq) --> BaS[tex]O_{4}[/tex] + 2HN[tex]O_{3}[/tex](aq)
Mass of [tex]Ba(NO_{3)3[/tex] = wt% * mass of the solid
= 23.2 x 4.35 / 100
= 1.01 g
Number of moles of [tex]Ba(NO_{3)2[/tex] = Mass / Molar Mass
Molar mass of [tex]Ba(NO_{3})2[/tex] = 137 + (14 + (16 * 3))*2
= 261 g/mol
= 1.01 / 261
= 0.00387
Since the chemical reaction occurs between one mole of Ba(NO3)2 and one mole of [tex]H_{2[/tex] [tex]SO_{4}[/tex]. Therefore, number of moles of [tex]H_{2}[/tex] [tex]SO_{4}[/tex] = 0.00387 moles. The formula is
Volume = number of moles/molar concentration
= 0.00387/3
= 0.00129 L
V2 = 1.29 mL
Hence, 1.29 milliliters of 3.00 m [tex]H_{2}[/tex] [tex]SO_{4}[/tex] (Sulfuric acid) are required to react with 4.35 g of a solid mixture containing 23.2 % (w/w) Ba(NO3)2.
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The complete question is:
How many milliliters of 3.00 m H2SO4 are required to react with 4.35 g of a solid mixture containing 23.2 % (w:w) Ba(NO3)2? (i.e. in the original sample, there are 23.2 g of barium nitrate per 100 g of total sample).
