If $K_1$ and $K_2$ are respective equilibrium constants for the two reactions:
$XeF_{6(g)} + H_2O_{(g)} \rightleftharpoons XeOF_{4(g)} + 2HF_{(g)}$
$XeO_{4(g)} + XeF_{6(g)} \rightleftharpoons XeOF_{4(g)} + XeO_3F_{2(g)}$
The equilibrium constant for the reaction $XeO_{4(g)} + 2HF_{(g)} \rightleftharpoons XeO_3F_{2(g)} + H_2O_{(g)}$ will be:

• A
  $\frac{K_1}{K_2^2}$
• B
  $K_1K_2$
• C
  $\frac{K_1}{K_2}$
• D
  $\frac{K_2}{K_1}$