Question

Consider:

2Al (s) + 3Cl2 (g) ----> 2AlCl3 (s) ΔH=-1390.81 kJ

Which of the following correctly describes he magnitude and direction of the enthalpy change when 10.0g of AlCl3 forms?

a. 104.4 kJ, exothermic
b. 52.2 kJ, exothermic
c. 26.1 kJ, exothermic
d. 52.2 kJ, endothermic
e. 26.1 kJ, endothermic

Answer 1

The answer is B.

Step-by-step explanation:

1 mole of AlCl3 = 27 + 3*35.5 grams = 133.5 grams

1 mole / 133.5 = x / 10 Cross multiply

133.5 x = 1 * 10

x = 10 / 133.5

x = 0.0749 moles

Delta H when written the way it is, is on the left side and since it is negative, the reaction must give away 1390.81 kj. The reaction is exothermic.

So now you get the answer between AB and C by setting up a ratio.

2/0.0749 = -1390.81 / x Cross multiply

2x = 0.0749 * -1390.81

2x =- 104.18 Divide both sides by 2

2x//2 = -104.18 / 2

x = -52.09

B