How many moles of $Al$ $"Al"$ are needed to form $3.7$ $3.7$ moles of ${Al}_{2} O_{3}$ $"Al"_2"O"_3$ ?

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How many moles of $Al$ $"Al"$ are needed to form $3.7$ $3.7$ moles of ${Al}_{2} O_{3}$ $"Al"_2"O"_3$ ?

$4 Al + 3 O_{2} \to 2 {Al}_{2} O_{3}$ $4"Al" + 3"O"_2 -> 2"Al"_2"O"_3$

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Answer 1 · 2018-02-20T03:06:50

$7.4 mol Al$ $"7.4 mol Al"$

Explanation:

The balanced chemical equation for it is

$4 A l + 3 O_{2} \to 2 A l_{2} O_{3}$ $4Al + 3O_2 -> 2Al_2O_3$

You have $3.7$ $3.7$ moles of $A l_{2} O_{3}$ $Al_2O_3$ . When you balanced the equation, you got $2$ $2$ moles of $A l_{2} O_{3}$ $"Al_2O_3$ and $4$ $4$ moles of $A l$ $Al$ , so

"4 mol $A l$ $Al$ gives $2$ $2$ mol $A l_{2} O_{3}$ $Al_2O_3$

After you cancel out the units of measurement to check the order of the conversion, multiply $3.7$ $3.7$ with $4$ $4$ and then divide the answer by $2$ $2$ . You get $7.4$ $7.4$ moles of $A l$ $Al$ .

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How many moles of $Al$ $"Al"$ are needed to form $3.7$ $3.7$ moles of ${Al}_{2} O_{3}$ $"Al"_2"O"_3$ ?

$4 Al + 3 O_{2} \to 2 {Al}_{2} O_{3}$ $4"Al" + 3"O"_2 -> 2"Al"_2"O"_3$

1 Answer

Explanation:

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How many moles of "Al"Al"Al" are needed to form 3.73.73.7 moles of "Al"_2"O"_3Al2O3"Al"_2"O"_3?

4"Al" + 3"O"_2 -> 2"Al"_2"O"_34Al+3O2→2Al2O34"Al" + 3"O"_2 -> 2"Al"_2"O"_3

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Explanation:

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How many moles of $Al$ $"Al"$ are needed to form $3.7$ $3.7$ moles of ${Al}_{2} O_{3}$ $"Al"_2"O"_3$ ?

$4 Al + 3 O_{2} \to 2 {Al}_{2} O_{3}$ $4"Al" + 3"O"_2 -> 2"Al"_2"O"_3$