Question

Consider the following reaction: 2Mg(s)+O2(g)-->2MgO(s) delta H=-1204kJ

Answer 1

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!