Respuesta :
Answer:
Maximum mass of [tex]I_2[/tex] produces=50.8g;
Explanation:
Firstly balance the chemical reaction,
[tex]I_2 O_5+5CO=5CO_2+I_2[/tex]
Molecular weight of [tex]I_2 O_5[/tex]=334g/mole;
Molecular weight of CO =28g/mole;
Number of mole of [tex]I_2 O_5[/tex] given=80g/(334g/mole)=0.239mole;
Number of mole of CO given=28g/(28g/mole)=1 mole
Moles are:
Number of mole of [tex]I_2 O_5[/tex]=0.239
Number of mole of CO=1;
From the balanced equation,
1 mole of [tex]I_2 O_5[/tex] reacts with 5 moles of CO;
Hence;
0.239 mole of [tex]I_2 O_5[/tex] will react with 5*0.239 moles of CO means
Moles of CO that reacts with [tex]I_2 O_5[/tex] =1.2 mole but we have given 1 mole therefore
[tex]I_2 O_5[/tex] 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[tex]I_2[/tex] of [tex]I_2[/tex];
1 mole of CO produces 1/5 mole of [tex]I_2[/tex];
Maximum mass of [tex]I_2[/tex] produces=mole*molecular weight;
Molecular weight of [tex]I_2[/tex]=254g/mole;
Maximum mass of produces [tex]I_2[/tex]= (1/5)*254=50.8g;
Maximum mass of [tex]I_2[/tex] produces=50.8g;
