Question

Consider the reaction I2O5(g) 5 CO(g) -------> 5 CO2(g) I2(g) a) 80.0 grams of iodine(V) oxide, I2O5, reacts with 28.0 grams of carbon monoxide, CO. Determine the mass of iodine I2, which could be produced

Answer 1

Answer:

Maximum mass of $I_2$ produces=50.8g;

Explanation:

Firstly balance the chemical reaction,

$I_2 O_5+5CO=5CO_2+I_2$

Molecular weight of $I_2 O_5$=334g/mole;  

Molecular weight of CO =28g/mole;

Number of mole of $I_2 O_5$ given=80g/(334g/mole)=0.239mole;

Number of mole of CO given=28g/(28g/mole)=1 mole  

Moles are:

Number of mole of $I_2 O_5$=0.239

Number of mole of CO=1;

From the balanced equation,

1 mole of $I_2 O_5$ reacts with 5 moles of CO;

Hence;

0.239 mole of $I_2 O_5$  will react with 5*0.239 moles of CO means

Moles of CO that reacts with $I_2 O_5$  =1.2 mole but we have given 1 mole therefore

$I_2 O_5$ Will be excess reagent and CO will be the limitting reagent  

And mass of product depends on limiting reagent i.e. mass of CO.

5 mole of CO produces 1 mole$I_2$ of $I_2$;

1 mole of CO  produces 1/5 mole of $I_2$;

Maximum mass of $I_2$  produces=mole*molecular weight;

Molecular weight of $I_2$=254g/mole;

Maximum mass of   produces $I_2$= (1/5)*254=50.8g;

Maximum mass of $I_2$ produces=50.8g;