Answer:
The answer to your question is: 70.7 %
Explanation:
Equation
Xe + 2F₂ ⇒ XeF₄
limiting reactant = Xe
Xe is the limiting reactant because the ratio is:
theoretical = 131/ 76 = 1.72 g
experimental ratio = 130/100 = 1.3 the amount of F increased.
131.3 g of Xe ------------------ 207 g of XeF₄
130 g of Xe ------------------- x
x = (130 x 207) / 131.3
x = 205 g of XeF₄
% yield = 145 / 205 x 100
% yield = 70.7