Using the reaction, $4 A l + 3 O_{2} \to 2 A l_{2} O_{3}$ $4Al + 3O_2 -> 2Al_2O_3$ , how do you identify the limiting reactant in each of the following: 0.25 mol Al and .40mol $O_{2}$ $O_2$ and 58.5g Al and 98.0g $O_{2}$ $O_2$ ?

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Using the reaction, $4 A l + 3 O_{2} \to 2 A l_{2} O_{3}$ $4Al + 3O_2 -> 2Al_2O_3$ , how do you identify the limiting reactant in each of the following: 0.25 mol Al and .40mol $O_{2}$ $O_2$ and 58.5g Al and 98.0g $O_{2}$ $O_2$ ?

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Answer 1 · 2017-07-19T05:47:23

Al

Explanation:

The limiting reactant is the reactant that is present in less amount. 4 mol of Al need to react completely of 3 mol of Oxygen (according with the reaction you have written). Making a proportion, you have 0,25 mol of Al so you need 0,25 x 3/4 = 0,1875 mol di Oxygen. You have 0,4 of that, therefore the limiting reactant is Al. in the second case you have $\frac{58, 5 g}{27 \frac{g}{m o l}} = 2, 17 m o l A l$ $(58,5g)/(27g/(mol)) =2,17 mol Al$ and $\frac{98 g}{32 \frac{g}{m o l}} = 3, 06 m o l O_{2}$ $(98g)/ (32g/(mol))= 3,06 mol O_2$ . Making a proportion, for 2,17 mol of Al you need 1,62 mol of Oxygen < 3,06. The limiting reactant is Al again

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Using the reaction, $4 A l + 3 O_{2} \to 2 A l_{2} O_{3}$ $4Al + 3O_2 -> 2Al_2O_3$ , how do you identify the limiting reactant in each of the following: 0.25 mol Al and .40mol $O_{2}$ $O_2$ and 58.5g Al and 98.0g $O_{2}$ $O_2$ ?

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Explanation:

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Using the reaction, 4Al+3O2→2Al2O34Al + 3O_2 -> 2Al_2O_3, how do you identify the limiting reactant in each of the following: 0.25 mol Al and .40mol O2O_2 and 58.5g Al and 98.0g O2O_2?

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Explanation:

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Using the reaction, $4 A l + 3 O_{2} \to 2 A l_{2} O_{3}$ $4Al + 3O_2 -> 2Al_2O_3$ , how do you identify the limiting reactant in each of the following: 0.25 mol Al and .40mol $O_{2}$ $O_2$ and 58.5g Al and 98.0g $O_{2}$ $O_2$ ?