Question

Oxygen difluoride can be prepared by bubbling gaseous fluorine into a 0.5 M solution of NaOH 2F2g) + 2 NaOH (a)---OF2(g) + 2 NaF (aq) + H2O(l) OF2 can be used to prepare compounds such as O2AsF6, containing the oxygen cation, O2+, by the following reaction: 4 OF@ + 2 AsF2----- 2 O2AsF6 + 3 F2 If 14 g of F2 is bubbled through 650 mL of 0.500M of NaOH to prepare OF2 with a 78% yield, how much O2AsF6 can be prepared?

Answer 1

The moles of the reactants are calculated:

n NaOH = M * V = 0.5 M * 0.65 L = 0.325 mol

n F2 = g / MM = 14 g / 38 g / mol = 0.368 mol

According to the reaction, the limit reagent is NaOH, the amount of OF2 produced is calculated:

n OF2 = 0.325 mol NaOH * (1 mol OF2 / 2 mol NaOH) * (78/100) = 0.127 mol

The mass of O2AsF6 that can be produced with this amount of OF2 is calculated:

g O2AsF6 = 0.127 mol OF2 * (2 mol OsAsF6 / 4 mol OF2) * (221 g / 1 mol) = 14.03 g

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