The tablets were crushed, and 4.9993 g of the powder was transferred to a beaker and reacted with HCl. After filtration, the filtrate was transferred to a 100-mL volumetric flask and diluted with water. 20.00 mL of this stock solution were combined with 0.2 M Na3PO4. The resulting precipitate weighed 0.3451 g after drying. Calculate the moles of BiPO4 precipitated, the moles of Bi3 in the stock solution, and the mass of BSS per tablet.
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Ответ:
Explanation:
From the information given:
Mass of BiPO₄ = 0.3451 g
Number of moles of BiPO₄ =![0.3451 \ g \ BiPO_4 \times \dfrac{1 \ mol \ BiPO_4}{303.95 \ g \ BiPO_4}](/tpl/images/1247/2786/1cc43.png)
The number of moles of Bi³⁺ in 20.00 mL is:![= 0.001135 \ mol \ BiPO_4 \times \dfrac{1 \ mol \ of \ Bi^{3+}}{1 \ mol \ BiPO_4}](/tpl/images/1247/2786/0db4f.png)
= 0.001135 mol of Bi³⁺
The number of moles of Bi³⁺ in 100 mL stock solution
Mass of BSS in 4.9993 g tablets
m = 2.055 g BSS
Mass of BSS in 5.0103 g (5 tables)
= 2.06 g
∴
The mass of BSS per tablet is![=\dfrac{2.06 \ g}{5 \ tablet}](/tpl/images/1247/2786/95671.png)
= 0.412 g BSS/ tablet
Ответ: