2XeF6 + SiO2 --> 2XeOF4 + SiF4 Consider the above chemical reaction. If 122.6 g of XeF6 reacts with 60 g of SiO2 to form the products. Select the limiting reagent and amount of SiF4 formed: (XeF6 245.3 amu, SiO2 = 60 amu, SiF4 = 104 amu)

Explanation

To solve this problem, first calculate the moles of each reactant. For XeF6, with a molar mass of 245.3 g/mol, 122.6 g is equivalent to 0.4986 moles. For SiO2, with a molar mass of 60 g/mol, 60 g is equivalent to 1 mole. The balanced equation shows that 2 moles of XeF6 react with 1 mole of SiO2. Since we have fewer moles of XeF6 compared to the stoichiometric requirement, XeF6 is the limiting reagent.Knowing this, we can calculate the amount of product formed. According to the stoichiometry of the reaction, 2 moles of XeF6 produce 1 mole of SiF4. Therefore, 0.4986 moles of XeF6 will produce 0.2493 moles of SiF4. The mass of SiF4 produced is 0.2493 moles multiplied by its molar mass of 104 g/mol, resulting in 25.5 g of SiF4.Therefore, the correct answer is that XeF6 is the limiting reagent, and 25.5 g of SiF4 is formed.The other options are incorrect because they either misidentify the limiting reagent or incorrectly calculate the amount of SiF4 formed.