How do you convert 9.45 moles g of $A l_{2} O_{3}$ $Al_2O_3$ into grams of $A l_{2} O_{3}$ $Al_2O_3$ ?

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How do you convert 9.45 moles g of $A l_{2} O_{3}$ $Al_2O_3$ into grams of $A l_{2} O_{3}$ $Al_2O_3$ ?

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Answer 1 · 2015-12-09T04:54:48

$963.9 g$ $963.9g$

Explanation:

consider the relationship between the number of moles, molar mass and the mass of a compound,

$n = \frac{m}{M}$ $n= m/M$

where n is the number of moles, m is the mass of the compound and M is the molar mass of the compound.

you just have to multiply the number of moles (9.45 moles) by the molar mass of $A l_{2} O_{3}$ $Al_"2"O_"3"$ which is $27 \cdot 2 + 16 \cdot 3 = 102$ $27*2+16*3 = 102$

so,
$m = 9.45 m o l \cdot 102 g m o l^{-1}$ $m = 9.45 mol * 102gmol ^"-1"$

$m = 963.9 g$ $m = 963.9g$

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How do you convert 9.45 moles g of $A l_{2} O_{3}$ $Al_2O_3$ into grams of $A l_{2} O_{3}$ $Al_2O_3$ ?

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How do you convert 9.45 moles g of Al2O3Al_2O_3 into grams of Al2O3Al_2O_3?

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How do you convert 9.45 moles g of $A l_{2} O_{3}$ $Al_2O_3$ into grams of $A l_{2} O_{3}$ $Al_2O_3$ ?