Part VII a roduce NO n, 57. To determine the percentage of water in a hydra r 20

Question

Part VII a roduce NO n, 57. To determine the percentage of water in a hydra r 20 alt, a student heated a 2.456 g sample of the salt fo minutes. The sample weighed 2.024 g after cooling. The sample was heated for another 10 minutes, and weighed 1.999 g after being cooled to room temperature. The sample was heated for another 10 minutes, and weighed 1.992 g after being cooled to room temperature. What should the student do next? a) Use the smallest mass value to calculate the percentage of water in the hydrated salt b) Reheat the sample for ten minutes and weigh it again while it is hot; repeat until a constant mass is obtained. c) Reheat the sample for ten minutes and weigh it again after cooling; repeat until a constant mass is obtained. d) Use the average mass of the two weighings to calculate the percentage of water in the hydrated salt. Che Part VII b 58. What is the maximum number of moles of Al,O that can be produced by the reaction of 0.60 mol Al with 0.3 mol O,? a) 0.10 b) 0.20 c)0.30 d) 0.40

Explanation / Answer

Ans. #57. Correct option- C. Reheat the sample for 10 minutes and weigh it again after cooling; repeat until a constant mass is obtained.

# Hot samples may absorb moisture from air which may give apparently increased dried mass of the sample. Therefore, the sample must be cooling to room temperature in a desiccator before cooling.

Since the third heating gives a lower mass (indicating loss of water of hydration) from sample, the subsequent heating and cooling events must be carried out until a constant mass is obtained.

The lowest constant, dry mass of sample is taken as dehydrated mass of sample and used to determine % water in hydrated salt.

#58. Following stoichiometry, 2 mol Al reacts with 3 mol O to produce 1 mol Al2O3.

So,

            Theoretical molar ratio of reactants:          Al : O = 2 : 3 = 1 : 1.5

# Moles of Al taken = 0.60

Moles of O taken = 2 x moles of O2 = 2 x 0.30 mol = 0.60 mol

Theoretical molar ratio of reactants:          Al : O = 0.60 : 0.60 = 1 : 1

Comparing the theoretical and experimental molar ration, the experimental moles of O is less than its theoretical value of 1.5 while keeping those of Al constant at 1.0 mol.

Therefore, O is the liming reactant.

# The formation of product follows the stoichiometry of limiting reactant.

Following stoichiometry, 3 mol O forms 1 mol Al2O­3.

So,

            Moles of Al2O3 formed = (1 / 3) x Moles of O = (1/3) x 0.60 mol = 0.20 mol

So, correct option is- b. 0.20

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---