Question

How many grams of O_2 will be produced if 23.2 g of XeF_2 reacts with excess water?

Answer 1

0.0685 moles of O2 will be produced

Step-by-step explanation:

To find the number of grams of O2 produced, we need to first determine the number of moles of XeF2 that react. From the balanced chemical equation, we can see that 2 moles of XeF2 react to produce 1 mole of O2. First, we need to convert the mass of XeF2 to moles using its molar mass (169.3 g/mol).

23.2 g XeF2 * (1 mol XeF2/169.3 g XeF2) = 0.1370 mol XeF2

Finally, we can use the mole ratio from the balanced equation to find the number of moles of O2 produced:

0.1370 mol XeF2 * (1 mol O2/2 moles XeF2) = 0.0685 mol O2

Therefore, 0.0685 moles of O2 will be produced.