Answer :
Final answer:
Using the Ideal Gas Law and stoichiometry, approximately 0.592 moles of oxygen form about 40.27 grams of Al2O3. Thus, none of the options (a, b, c, d) provided in the question are correct as the answer is around 40.27 g of Al2O3.
Explanation:
The question is asking about stoichiometry, the calculation of relative quantities of reactants and products in chemical reactions, specifically, how many grams of Al2O3 are formed from 15L of O2. To solve this, we'll use the Ideal Gas Law (PV = nRT) to figure out the moles of O2, and then use stoichiometry to determine the weight of Al2O3 created.
First, we convert the temperature to Kelvin by adding 273 to the 21°C to get 294K. Then, we convert the pressure from kPa to atm by dividing 97.5 kPa by 101.3 (1 atm = 101.3 kPa) to get approximately 0.963 atm. Plug the values into the Ideal Gas Law: (0.963 atm)(15 L) = n(0.0821 L·atm/mol·K)(294 K). Solve for n (moles of O2) and you get approximately 0.592 moles of O2.
In the reaction 4Al + 3O2 -> 2Al2O3, the stoichiometry is such that 3 moles of O2 gives 2 moles of Al2O3. So 0.592 moles of O2 will form (0.592*2)/3 = 0.395 moles Al2O3. The molar mass of Al2O3 is approximately 101.96 g/mol, so the mass of the Al2O3 formed will be 0.395 moles * 101.96 g/mol = 40.27 g
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