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The combustion of butane produces heat according to the equation 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l), ΔH°rxn= –5,314 kJ/mol. How many grams of CO2 are produced per 1.00 × 104 kJ of heat released?

Respuesta :

dsdrajlin

Answer:

665 g

Explanation:

Let's consider the following thermochemical equation.

2 C₄H₁₀(g) + 13 O₂(g) → 8 CO₂(g) + 10 H₂O(l), ΔH°rxn= –5,314 kJ/mol

According to this equation, 5,314 kJ are released per 8 moles of CO₂. The moles produced when 1.00 × 10⁴ kJ are released are:

-1.00 × 10⁴ kJ × (8 mol CO₂/-5,314 kJ) = 15.1 mol CO₂

The molar mass of CO₂ is 44.01 g/mol. The mass corresponding to 15.1 moles is:

15.1 mol × 44.01 g/mol = 665 g

pstnonsonjoku

The mass of carbon dioxide produced is 662 g.

We have the equation  of the reaction as follows;

2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l)

The we are told that the ΔH°rxn= –5,314 kJ/mol for the reaction shown above.

Now, from the balanced reaction equation;

2 moles of butane produces  –5,314 kJ of heat

x moles of butane produces 1.00 × 10^4 kJ

x moles = 2 moles ×1.00 × 10^4/ 5,314

x =3.76 moles

Since 2 moles of butane produces 8 moles of carbon dioxide

3.76 moles moles of butane produces  = 3.76 moles ×  8 moles /2 moles

= 15.04 moles

Mass of carbon dioxide =15.04 moles × 44 g/mol = 662 g

Learn more: https://brainly.com/question/14281129

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