Answer :
A mass of 4.9 x 10^-4 g of bismuth hydroxide is needed to create one dosage.
Equation
[tex]3Bi(OH)_{3} +3HC_{7}H_{5} O_{3} \longrightarrow Bi_{3} C_{21}H_{15} O_{12}+3H_{2} O[/tex]
m(Bi3C21H15O12)=0.600mg=0.0006g
n(Bi3C21H15O12)=m/Mr=5,52*10^{-7} mol
n([tex]Bi(OH)_{3}[/tex] theor)= 3n (Bi3C21H15O12) = 1.67 x 10^{-6} mol
n([tex]Bi(OH)_{3}[/tex],exp)= 1.67 x 10^{-6} mol / yield = 1.88 x 10^{-6} mol
m([tex]Bi(OH)_{3}[/tex]} = n x Mr = 4.9 x 10^{-4} g
A mass of 4.9 x 10^-4 g of bismuth hydroxide is needed to create one dosage.
What constitutes the majority of Pepto-Bismol?
A fragile, crystalline, white metal with a faint pink undertone, bismuth is also brittle. It can be used in a wide range of products, such as cosmetics, alloys, fire extinguishers, and weapons. It is most famous for being the major component of stomach discomfort medications like Pepto-Bismol. A mass of 4.9 x 10-4g of bismuth hydroxide is needed to create one dosage.
What does Pepto-Bismol actually do?
This drug is used to treat nausea, heartburn, and sporadic stomach distress. Additionally, it is employed to cure and lessen the effects of travelers' diarrhea.
learn more about Pepto-Bismol here
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