# If 29.4 grams of hydrogen and 3701.3 grams of iodine combine to form hydrogen iodide, how many grams of hydrogen iodide must form?

Apr 5, 2016

$3.730 k g$

#### Explanation:

First write down a balanced chemical equation for the reaction :

${H}_{2} + {I}_{2} \to 2 H I$

This represents the mole ratio in which chemicals combine.

Now calculate the number of moles of each reactant present :

${n}_{{H}_{2}} = \frac{m}{M} _ r = \frac{29.4}{2} = 14.7 m o l$

${n}_{{I}_{2}} = \frac{m}{M} _ r = \frac{3701.3}{253.8} = 14.583 m o l$

Since the balanced chemical equation illustrates that the reactants combine in a $1 : 1$ ratio, it implies that iodine is the limiting reagent and decides how much product is formed. Hydrogen is in excess.

So hence, since $1 m o l {I}_{2}$ produces $2 m o l H I$, it follows that $14.583 m o l {I}_{2}$ will produce $29.166 m o l H I$

This corresponds to a mass of product of

$m = n \times {M}_{r}$

$= 29.166 \times 127.9$

$= 3730.33 g$