Question

C2H6O(g) + 3O2(g) 2CO2(g) + 3H2O(g) Inside the piston of an automobile engine, 0.461 g of ethanol (C2H6O) gas reacts with 0.640 grams of oxygen gas to produce carbon dioxide and water, according to the balanced equation shown above. What is the limiting reagent?

Answer 1

Answer:

Oxygen O₂ will be the limiting reagent.

Explanation:

C₂H₆O(g) + 3 O₂(g) ⇒ 2 CO₂(g) + 3 H₂O(g)

First of all you must know by stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction) how much mass of each compound reacts. First of all, being:

• C: 12 g/mole
• H: 1 g/mole
• O: 16 g/mole

the molar mass of each reagent is:

• C₂H₆O: 2*12 g/mole + 6* 1 g/mole + 16 g/mole= 46 g/mole
• O₂: 2*16 g/mole= 32 g/mole

Then, since 1 mol of C₂H₆O and 3 moles of O₂ react by stoichiometry, the amount of mass that reacts is:

• C₂H₆O: 1 mole* 46 g/mole= 46 g
• O₂: 3* 32 g/mole= 96 g

Now you apply a rule of three as follows: if 46 g of C₂H₆O reacts with 96 g of O₂ by stoichiometry, 0.461 g of C₂H₆O with how much mass of O₂ will they react?

$mass of O_{2} =\frac{0.461 grams of C_{2}H_{6} O*96 grams ofO_{2} }{46 grams of C_{2}H_{6} O}$

mass of O₂= 0.962 grams

But 0.962 grams of O₂ are not available, 0.640 grams are available. Since you have less mass than you need to react with 0.461 g of C₂H₆O, oxygen O₂ will be the limiting reagent.

Answer 2

Answer:

Oxygen is the limiting reactant.

Explanation:

Hello,

In this case, given the reaction:

$C_2H_6O(g) + 3O_2(g)\rightarrow 2CO_2(g) + 3H_2O(g)$

Hence, given the masses of both ethanol and oxygen, we are able to compute the available moles ethanol by:

$n_{C_2H_6O}^{available}=0.461g*\frac{1mol}{46g}=0.01mol C_2H_6O$

Next, we compute the moles of ethanol that react with the 0.640 grams of oxygen considering their 1:3 molar ratio in the chemical reaction:

$n_{C_2H_6O}^{consumed\ by\ O_2}=0.64gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6O}{3molO_2} =0.0067molC_2H_6O$

In such a way, since there are 0.01 available moles of ethanol but just 0.0067 moles are reacting, we evidence ethanol is in excess, therefore the oxygen is the limiting reactant.

Best regards.