In the presence of excess oxygen,methane gas burns in a constant-pressure system to yield carbon dioxide and water:
CH₄(g) +2O₂ (g) → CO₂(g) + 2H₂O (l)
H = -890.0 kJ
Calculate the value of q (kJ) in this exothermic reaction when 1.90 g of methane is combusted at constant pressure.

Question

In the presence of excess oxygen,methane gas burns in a constant-pressure system to yield carbon dioxide and water:
CH₄(g) +2O₂ (g) → CO₂(g) + 2H₂O (l)

In the presence of excess oxygen,methane gas burns in a constant-pressure system to yield carbon dioxide and water: CH<sub>4</sub>(g) +2O<sub>2</sub> (g) → CO<sub>2</sub>(g) + 2H<sub>2</sub>O (l) H = -890.0 kJ Calculate the value of q (kJ) in this exothermic reaction when 1.90 g of methane is combusted at constant pressure. A) -105.7 kJ B) 0.0342 kJ C) -0.0095 kJ D) 29.3 kJ E) -1.06 × 10<sup>5</sup> kJ

H = -890.0 kJ
Calculate the value of q (kJ) in this exothermic reaction when 1.90 g of methane is combusted at constant pressure.

Quizplus · Accepted Answer

To solve for q, we need to use the equation:

q = nΔH

where n is the number of moles and ΔH is the enthalpy change of the reaction.

First, we need to calculate the number of moles of methane:

n = m/M

where m is the mass and M is the molar mass of methane.

Methane (CH4) has a molar mass of 16.04 g/mol.

n = 1.90 g / 16.04 g/mol = 0.1185 mol

Now we can substitute into the equation for q:

q = nΔH

q = (0.1185 mol) (-890.0 kJ/mol)

q = -105.67 kJ

Therefore, the answer is A because q is negative and has a value of -105.7 kJ, indicating an exothermic reaction.

In the Presence of Excess Oxygen,methane Gas Burns in a Constant-Pressure