How many moles of hydrogen will be produced in this reaction when 34.5 moles of hydrogen iodide are decomposed?

17.25 moles of hydrogen

The equation for the reaction;

2HI (g) → H₂(g) + i₂ (g)

34.5 moles of Hydrogen iodide are decomposed;

The mole ratio of HI to Hydrogen = 2 : 1

Therefore;

moles of hydrogen = (34.5/2) ×1

                                = 17.25 moles

Therefore; 17.25 moles of hydrogen will be produced when 34.5 moles of Hydrogen iodide are decomposed

anaphase.

hope this !

n+h→nh3

we must remember that nitrogen and hydrogen are both diatomic molecules in their standard gas form. this adjusts the equation to

n2+h2→nh3

now we need to adjust coefficients in order to balance the atoms on each side of the equation. currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side.

we can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen.

n2+3h2→2nh3

this gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side.