A chemistry has 14.5 g of Al and 7.5 g of $O_{2}$ $O_2$ . What would the excess be for the following reaction: $4 A l + 3 O_{2} \to 2 A l_{2} O_{3}$ $4Al + 3O_2 -> 2Al_2O_3$ ?

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A chemistry has 14.5 g of Al and 7.5 g of $O_{2}$ $O_2$ . What would the excess be for the following reaction: $4 A l + 3 O_{2} \to 2 A l_{2} O_{3}$ $4Al + 3O_2 -> 2Al_2O_3$ ?

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Answer 1 · 2017-09-08T11:22:34

Al is in limiting
$O_{2}$ $O_2$ is in excess

Explanation:

To check excess and limiting reactants, the number of moles has to be found, and then divided by the coefficient.
For the sake of shortening the equations, let us say that:

$n = number of moles$ $n="number of moles"$
$m = mass$ $m="mass"$
$M = molar mass$ $M="molar mass"$

To find the number of moles, we need to use the formula:

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

The mass is given in the question:
$m_{Al} = 14.5 g$ $m_"Al"=14.5g$
$m_{Oxygen gas} = 7.5 g$ $m_"Oxygen gas"=7.5g$

Molar mass of Al and Oxygen can be found in periodic table,
$M_{Al} \approx 27 \frac{g}{m} o l$ $M_"Al"≈27g/mol$
$M_{Oxygen gas} = 2 \cdot M_{Oxygen atom} \approx 2 \cdot 16 = 32 g/mol$ $M_"Oxygen gas"= 2*M_"Oxygen atom"≈2* 16=32 "g/mol"$
(since oxygen gas ( $O_{2}$ $O_2$ ) has two atoms of Oxygen (O))

Now putting these values for the number of moles's formula:
For Al

$n = \frac{m}{M} = \frac{14.5}{27} = 0.537 m o l s$ $n=m/M=14.5/27=0.537 mols$

For $O_{2}$ $O_2$

$n = \frac{7.5}{32} = 0.234 m o l s$ $n=7.5/32=0.234 mols$

After getting the number of moles, we just still have to divide them by their respective coefficients, and comparing them.

For Al

$\frac{n}{4} = \frac{0.537}{4} = 0.134$ $n/4=0.537/4=0.134$

For $O_{2}$ $O_2$

$\frac{n}{2} = \frac{0.234}{2} = 0.17$ $n/2=0.234/2=0.17$

The reactant with the bigger value will be in excess and the reactant with the smaller value will be in limiting

$the value of Al (0.134) < the value of O_2 (0.17)$ $"the value of Al (0.134)"< "the value of O_2 (0.17)"$

Therefore, Al is in limiting (there is less of it and thus it is limiting the reaction) and $O_{2}$ $O_2$ is in excess.

Hope that this will help. :)

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A chemistry has 14.5 g of Al and 7.5 g of $O_{2}$ $O_2$ . What would the excess be for the following reaction: $4 A l + 3 O_{2} \to 2 A l_{2} O_{3}$ $4Al + 3O_2 -> 2Al_2O_3$ ?

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Explanation:

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A chemistry has 14.5 g of Al and 7.5 g of O_2O2O_2. What would the excess be for the following reaction: 4Al + 3O_2 -> 2Al_2O_34Al+3O2→2Al2O34Al + 3O_2 -> 2Al_2O_3?

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A chemistry has 14.5 g of Al and 7.5 g of $O_{2}$ $O_2$ . What would the excess be for the following reaction: $4 A l + 3 O_{2} \to 2 A l_{2} O_{3}$ $4Al + 3O_2 -> 2Al_2O_3$ ?