Answer to Question #328230 in Chemistry for alxn

Question

Using the chemical reaction, Xe(g)+F 2 (g) XeF 4 (g) , how many ml of xexor tetrafluoride will be produced from 2.65 mL of xenon and 1.74 mL of fluorine?

Answer 1

Xe_(g) + 2F_2(g) = XeF_4(g)

n = V / 22.4

n (Xe) = 0.00265 / 22.4 = 0.0001 mol

n (F₂) = 0.00174 / 22.4 = 0.00008 mol

According to the equation, n (XeF₄) = 1/2 x n (F₂) = n (Xe)

F₂ is the limiting reagent.

n (XeF₄) = 1/2 x n (F₂) = 1/2 x 0.00008 = 0.00004 mol

V (XeF₄) = 0.00004 x 22.4 = 0.00087 L = 0.87 ml