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REY [17]
3 years ago
11

Which of the following statements is true of combustion reactions?

Chemistry
2 answers:
enot [183]3 years ago
8 0

Gasoline contains C and H atoms. During combustion, the carbon (C) from the fuel combines with oxygen (O2) from the air to produce carbon dioxide (CO2).
2 C8H18 + 25 O2 → 16 CO2 + 18 H2O.

Combustion reactions release large amounts of heat. They have negative enthalpy. A negative enthalpy represents an exothermic reaction, releasing heat. This reaction is spontaneous and exothermic, since we can obtain energy from the reaction; the ΔG (free energy) is negative (So 1 is true).

ΔG < 0, so the free energy of the system decreases with the reaction. Remember that when there is a negative ΔG the reaction goes from higher free energy to lower free energy, like in this case.

ipn [44]3 years ago
4 0

Answer: Option (B) is the correct answer.

Explanation:

A chemical reaction in which a hydrocarbon combines with oxygen atom that is present in air to yield carbon dioxide and water along with the products.

For example, CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O

Here, methane undergoes combustion reaction to yield carbon dioxide and water as the products.

Hence, we can conclude that the statement water (H_{2}O) is always a product, is true of combustion reactions.

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If 16.29 grams of Na2SO4 is mixed with 3.697 grams of C and allowed to react according to the balanced equation: Na2SO4(aq) + 4
abruzzese [7]

<u>Answer:</u> The limiting reagent in the given chemical reaction is carbon metal.

<u>Explanation:</u>

Excess reagent is defined as the reagent which is present in large amount in a chemical reaction.

Limiting reagent is defined as the reagent which is present in small amount in a chemical reaction. Formation of product depends on the limiting reagent.

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

  • <u>For sodium sulfate:</u>

Given mass of sodium sulfate = 16.29 g

Molar mass of sodium sulfate = 142 g/mol

Putting values in equation 1, we get:

\text{Moles of sodium sulfate}=\frac{16.29g}{142g/mol}=0.115mol

  • <u>For carbon:</u>

Given mass of carbon = 3.697 g

Molar mass of carbon = 12 g/mol

Putting values in equation 1, we get:

\text{Moles of carbon}=\frac{3.697g}{12g/mol}=0.31mol

For the given chemical reaction:

Na_2SO_4(aq.)+4C(s)\rightarrow Na_2S(s)+4CO(g)

By Stoichiometry of the reaction:

4 moles of carbon reacts with 1 mole of sodium sulfate

So, 0.31 moles of carbon will react with = \frac{1}{4}\times 0.31=0.0775mol of sodium sulfate

As, given amount of sodium sulfate is more than the required amount. So, it is considered as an excess reagent.

Thus, carbon metal is considered as a limiting reagent because it limits the formation of product.

Hence, the limiting reagent in the given chemical reaction is carbon metal.

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