How many mol of xenon hexafluoride are produced from 4.01 x 10^2 mol of fluorine (F2)?
Xe + 3F2 -> XeF6
A. 1.79 x 10^1 mol XeF6
B. 2.41 x 10^26 mol XeF6
C. 1.20 x 10^3 mol XeF6
D. 1.34 x 10^2 mol XeF6
Please help !
every 3 mols of F2 produces 1 mol of XeF6 according to your balanced equation
so
(4.01/3) * 10^2
So is the answer D?
Thanks!
You are welcome.
To determine how many moles of xenon hexafluoride (XeF6) are produced from a given amount of fluorine (F2), we need to use the stoichiometry of the balanced chemical equation.
First, let's examine the balanced equation:
Xe + 3F2 -> XeF6
From the equation, we can see that one mole of xenon (Xe) reacts with three moles of fluorine (F2) to produce one mole of xenon hexafluoride (XeF6).
Given that we have 4.01 × 10^2 moles of fluorine (F2), we can use this information to find the number of moles of xenon hexafluoride (XeF6).
Multiply the given amount of moles of fluorine (F2) by the stoichiometric ratio between F2 and XeF6 (3:1):
4.01 × 10^2 mol F2 × (1 mol XeF6 / 3 mol F2) = 1.34 × 10^2 mol XeF6
Therefore, the correct answer is option D: 1.34 × 10^2 mol XeF6.