Question #e273b

1 Answer
Dec 31, 2016

Answer:

The reaction formed 9.96 mol of #"HI"#.

Explanation:

You have the equation

#"H"_2 + "I"_2 → "2HI"; ΔH "= 264 kJ"#

If #Δ_"f"H# for #"HI"# is 26.5 kJ/mol, then

#"Amount of HI" =264 color(red)(cancel(color(black)("kJ"))) × "1 mol HI"/(26.5 color(red)(cancel(color(black)("kJ")))) = "9.96 mol HI"#