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A mixture of bases can sometimes be the active ingredient in antacid tablets. If 0.4826 g of a mixture of Al(OH)3and Mg(OH)2is neutralized with 17.30 mL of 1.000 M HNO3,what is the mass % ofAl(OH)3in the mixture?

Short Answer

Expert verified

The mass % of Al(OH)3in the mixture is 55.93%

Step by step solution

01

Write down the molecular equation

Balanced molecular equation:

Al(OH)3+Mg(OH)2(s)+5HNO3→Al(NO3)3+Mg(NO3)2+5H2O(l)

02

Calculate the mass of Al(OH)3

²Ô³Ü³¾²ú±ð°ù o´Ú″¾´Ç±ô±ð²õ o´Ú H±·°¿3=1″¾´Ç±ô/±ô×17.3³¾±ô×1l1000ml=0.0173mole

³¾´Ç±ô±ð â¶Ä‰r²¹³Ù¾±´Ç = 1 â¶Ä³¾´Ç±ô±ð´¡±ô(OH)35″¾´Ç±ô±ð²õ±á±·°¿3=15

Number of moles of Al(OH)3 that reacted with 0.0173″¾´Ç±ô±ð²õof HNO3is calculated below.

Number of″¾´Ç±ô±ð²õof Al(OH)3=0.0173×15=0.00346

Mass â¶Ä‰of 1 mole Al(OH)3=78.01gMass â¶Ä‰of â¶Ä‰0.00346 mole â¶Ä‰Al(OH)3=78.01g×0.00346=0.2699g

03

Determine the mass % of Al(OH)3

The mass percent of Al(OH)3:

mass%ofAl(OH)3=massofAl(OH)3totalmassofmixtureofAl(OH)3andMg(OH)2×100%=0.2699gmAl(OH)30.4826gmx100%=55.93%

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