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Kryger [21]
3 years ago
11

The limiting reactant is the chemical substance that determines the amount of product(s) that can ultimately be formed in a reac

tion. During the reaction, the limiting reactant is completely consumed or used up, and therefore, causes the reaction to stop. The limiting reactant can be identified through stoichiometric calculations. After comparing the results, the reactant that produces the smaller mass of product is identified as the limiting reactant. Determine the excess reactant and calculate the mass of the remaining excess reactant after 20.0 grams of Al and 10.0 grams of O2 react. 4 Al(s) + 3 O2(g) → 2 Al2O3(s) For the calculations in this module, the molar mass of an element will be rounded to the hundredths place (0.01 g).
Chemistry
1 answer:
Tamiku [17]3 years ago
5 0

Answer:

The reactant in excess is Al (s)

The mass of remaining excess reactant is 8.76 g.

Explanation:

This is the reaction:

4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)

4 moles of solid aluminun reacts with 3 moles of oxygen, to make 2 moles of alumina.

Molar mass Al: 26.98 g/m

Molar mass O₂ : 32.00 g/m

Mass / Molar mass : Moles

20 g Al / 26.98 g/m = 0.741 moles

10 g O₂ / 32.00 g/m =  0.312 moles

Let's find out the limiting reactant and the reactant in excess by rules of three:

4 moles of Al reacts with 3 moles of O₂

0.741 moles of Al reacts with  (0.741 .3)/4 = 0.555 moles

Now we know that are needed 0.555 moles of O₂ to consume all the Al but we only have 0.312 moles of O₂.

The O₂ is the limiting reactant, so the Al is the reactant in excess.

Let's make another rule of three

3 moles of O₂ reacts with 4 moles of Al

0.312 of O₂ reacts with  (0.312  .4)/3 = 0.416

We need 0.416 moles to consume all the O₂ and we have 0.741 moles of Al. There's still Al to consume, that's why the Al is the reactant in excess.

When we use 0.416 moles of 0.741 moles of Al, 0.325 moles are remaining to consume all the Al.

Let's calculate the mass

0.325 moles . 26.98 g/m = 8.76 grams

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If 5 grams of co and 5 grams of o2 are combined according to the reaction 2co o2 --> 2co2, which is the limiting reagent?
IrinaK [193]

When 5 grams of CO and 5grams of O₂ is combined according to the reaction 2CO + O₂ ----------> 2CO₂ then Limiting reagent is CO.

Limiting reagent is the reactant that get used up in the reaction first.

According to the given reaction:

2CO + O₂ ----------> 2CO₂

5g       5g

∴ Molar mass of CO = Molar mass of C + Molar mass of O

⇒ Molar mass of CO = 14 + 16

⇒ Molar mass of CO = 28g

∴ Molar mass of O₂ = 16(2) = 32g

∴ Molar mass of CO₂ = Molar mass of C + 2(Molar mass of O)

⇒ Molar mass of CO₂ = 14 + 2(16)

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Let's find out the moles of CO and O₂

∴ Moles = Given mass / Molar mass

⇒ moles of CO = 5/28 = 0.17

⇒ moles of O₂ = 5/32 = 0.15

For finding out the Limiting Reagent, we will divide the number of moles with the stiochiometry of the given reaction.

⇒ For CO = Moles/ stiochiometry = 0.17/2 = 0.085

⇒ For O₂ = Moles/ stiochiometry = 0.15/1 = 0.15

Since, the ratio of number of moles with the stiochiometry is less for CO hence it is the Limiting reagent, i.e. it will get used up in the reaction first.

Hence, the Limiting reagent  for the reaction is CO.

Learn more about Limiting reagent here, brainly.com/question/11848702

#SPJ4

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