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?

Md Ariful Islam

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Md Ariful Islam

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First write down a balanced chemical equation for the reaction :

$H_2+I_2->2HI$

This represents ratio in which chemicals combine.

Now calculate the number of moles of each reactant present :

$n_(H_2)=m/M_r=29.4/2=14.7mol$

$n_(I_2)=m/M_r=3701.3/253.8=14.583mol$

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

So hence, since $1mol I_2$ produces $2molHI$, it follows that $14.583molI_2$ will produce $29.166molHI$

This corresponds to a mass of product of

$m=nxxM_r$

$=29.166xx127.9$

$=3730.33g$

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