Consider the following reaction: 2Mg(s)+O2(g)-->2MgO(s) delta H=-1204kJa) Is this reaction exothermic or endothermic? exothermic endothermic
b) Calculate the amount of heat transferred when 3.55 g of Mg(s) reacts at constant pressure. delta H = -87.9 kJ
c) How many grams of MgO are produced during an enthalpy change of -238 kJ ? m = ?g
d) How many kilojoules of heat are absorbed when 40.5 g of MgO(s) is decomposed into Mg(s) and O2(g) at constant pressure? delta H =? kJ

Answers

Answer 1
Answer:

Answer:

a. The reaction is exothermic.

b. -87,9 kJ

c. 9,60g of Mg(s)

d. 602kJ are absorbed

Explanation:

Based on the reaction:

2Mg(s) + O₂(g) → 2MgO(s) ΔH = -1204kJ

a. The reaction is exothermic. Because ΔH<0. That means the reaction produces heat when occurs

b. 3,55g of Mg(s) are:

3,55g Mg × ( 1mol / 24,305g) = 0,146 moles of Mg(s)

As 2 moles of Mg(s) produce -1204 kJ of heat:

0,146 moles of Mg(s) × ( -1204kJ / 2mol Mg) =  -87,9 kJ

c. If -238 kJ of heat were transferred. The moles of Mg(s) that react must be:

-238kJ × ( 2mol Mg / -1204kJ) = 0,395 moles of Mg(s). In grams:

0,395 moles × ( 24,305g / 1mol Mg) = 9,60g of Mg(s)

d. The reverse reaction is:

2MgO(s) → 2Mg(s) + O₂(g)  ΔH = +1204kJ

40,5g of MgO(s) are:

40,5g MgO × ( 1mol MgO / 40,3044g) = 1,00 moles of MgO(s)

As 2 moles of MgO absorbe 1204kJ of energy:

1,00 moles of MgO(s) × ( +1204 kJ / 2mol MgO) = 602kJ are absorbed

I hope it helps!

Answer 2
Answer:

Final answer:

This response explains whether the reaction is exothermic or endothermic, calculates the amount of heat transferred in different scenarios, and determines the mass of MgO produced during a given enthalpy change. It also calculates the amount of heat absorbed during the decomposition of MgO.

Explanation:

a) This reaction is exothermic because the enthalpy change, represented by delta H, is a negative value (-1204 kJ).

b) To calculate the amount of heat transferred when 3.55 g of Mg(s) reacts, we can use the equation q = m * delta H. Plugging in the given values, we get q = (3.55 g) * (-87.9 kJ/1 mol) / (24.31 g/mol) = -15.98 kJ.

c) To determine the mass of MgO produced during an enthalpy change of -238 kJ, we can use the equation q = m * delta H and solve for m. Plugging in the given values, we get m = (-238 kJ) / (-1204 kJ/mol) * (40.31 g/mol) = 7.90 g.

d) To find the amount of heat absorbed when 40.5 g of MgO(s) is decomposed, we can again use the equation q = m * delta H. Plugging in the given values, we get q = (40.5 g) * (-1204 kJ/mol) / (40.31 g/mol) = -1209 kJ.

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Hello,

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