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Magnesium exists in nature as three stable isotopes. Suppose you determine that 9.87 moles of natural magnesium are required if you wish to isolate 1.00 mole of pure Mg-25 (24.9858 amu). The most abundant isotope is Mg-24 with a mass of 23.9850 amu. Determine how many grams of the third isotope, Mg-26 (25.9826 amu), can be isolated from the 9.87 mole sample of natural magnesium.

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Answer:  28.2 g of ²⁶Mg can be isolated from the 9.87 mole sample of natural magnesium

Explanation:

we know that natural abundance of ²⁶Mg is 11.01 %

therefore moles of ²⁶Mg isolated from 9.87 mole sample of natural magnesium will be :

( 11.01 x 9.87 mol) / 100  = 1.09 mol

so

mass of ²⁶Mg = mol x molar mass

⇒ 1.09 mol x 25.9826 g/mol

= 28.2 g

Therefore 28.2 g of ²⁶Mg can be isolated from the 9.87 mole sample of natural magnesium

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